15. Algorithm Analysis & Recursion.ipynb

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   "source": [
    "\n",
    "# <font color=red>**Algorithm Analysis & Recursion**</font>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The goal of algorithm analysis is:\n",
    "- To find a way to obtain the most efficient solution for each problem; and\n",
    "- More importantly, to confirm that it is indeed the most efficient. "
   ]
  },
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   "source": [
    "\n",
    "## <font color='blue'>Search problem in an array (finding a specific item). </font>\n",
    "- <font color=blue> Linear Search (order of growth is linear) </font>\n",
    "-  <font color=blue> Binary Search  (order of growth is logarithmic) are two methods of searching</font>"
   ]
  },
  {
   "cell_type": "markdown",
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    "id": "WRZQHTb5TGCz"
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   "source": [
    "## <font color=blue> **Linear Search** </font>\n",
    "\n",
    "<font>**Linear time – O(n)**</font>\n",
    "\n",
    "O(n) denotes that the output is proportional to the size of the input\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "2Kc-2qHRSa5W"
   },
   "source": [
    "## <font color=blue> Time Linear Search algorithm </font>"
   ]
  },
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     "text": [
      "Wall time: 14.1 s\n"
     ]
    }
   ],
   "source": [
    "%%time\n",
    "\n",
    "from time import sleep\n",
    "\n",
    "# Get Element by Linear Search\n",
    "def LinearSearch(lys, element):\n",
    "    for i in range (len(lys)):\n",
    "        sleep(1)\n",
    "        if lys[i] == element:\n",
    "            return i\n",
    "    return -1\n",
    "\n",
    "arr = [12,31,35,37,41,43,47,51,53,59,61,67,71,73,79,83,97]\n",
    "val = 73\n",
    "\n",
    "\n",
    "res = LinearSearch(arr, val)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "- To highlight the number of steps taken, wait for 1 sec is introduced before executing the next instance of the loop.\n",
    "- 14 sec is observed because it takes 14 seconds"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "iq1-VRm8SYmL"
   },
   "source": [
    "## **Binary Search**\n",
    "\n",
    "<font>**A little faster...**</font>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "G8eIoinNSa5a"
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   "source": [
    "## <font color=blue> Time Binary Search algorithm </font>"
   ]
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     "text": [
      "Wall time: 4.05 s\n"
     ]
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   "source": [
    "%%time\n",
    "\n",
    "from time import sleep\n",
    "\n",
    "# Get Element by Binary Search\n",
    "def BinarySearch(lys, val):\n",
    "    first = 0\n",
    "    last = len(lys)-1\n",
    "    index = -1\n",
    "    while (first <= last) and (index == -1):\n",
    "        sleep(1)\n",
    "        mid = (first+last)//2\n",
    "        if lys[mid] == val:\n",
    "            index = mid\n",
    "        else:\n",
    "            if val<lys[mid]:\n",
    "                last = mid -1\n",
    "            else:\n",
    "                first = mid +1\n",
    "    return index\n",
    "\n",
    "arr = [12,31,35,37,41,43,47,51,53,59,61,67,71,73,79,83,97]\n",
    "val = 73\n",
    "\n",
    "res = BinarySearch(arr, val)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Suppose we have a function f(x). Now, irrespective of what input we provide, the function should take approximately the same amount of time to run."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Deterministic algorithms**: the time taken for execution is constant even if the input is varied, keeping the input size constant. As the name ‘deterministic’ suggests, the runtime can be estimated as a function of its input size."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Algorithm types\n",
    "- Deterministic\n",
    "- Probabilistic"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
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     "text": [
      "142\n"
     ]
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   "source": [
    "# Deterministic algorithms\n",
    "# the time taken for execution is constant even if the input is varied, keeping the input size constant.\n",
    "\n",
    "def sum(list):\n",
    "    sum=0\n",
    "    for i in list:\n",
    "        sum=sum+i\n",
    "    return sum\n",
    "\n",
    "list=[1,6,77,5,4,5,5,34,5]\n",
    "print(sum(list))"
   ]
  },
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     "text": [
      "13\n"
     ]
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   "source": [
    "# Probabilistic algorithms\n",
    "\n",
    "#Linear Search\n",
    "def LinearSearch(lys, element):\n",
    "    for i in range (len(lys)):\n",
    "        if lys[i] == element:\n",
    "            return i\n",
    "    return -1\n",
    "\n",
    "arr = [12,31,35,37,41,43,47,51,53,59,61,67,71,73,79,83,97]\n",
    "val = 73\n",
    "\n",
    "\n",
    "print(LinearSearch(arr, val))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The problem with probabilistic algorithms is that their runtime is usually dependent on factors beyond the programmer's control, such as the user's input. So, in order to analyse such probabilistic algorithms, we need to come up with the concept of scenarios."
   ]
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   "source": [
    "---\n",
    "\n",
    "# **Scenarios**\n",
    "\n",
    "\n",
    "\n",
    "**We have three cases for every algorithm**\n",
    "* Best Case denoted by omega symbol\n",
    "* Average Case denoted by theta symol\n",
    "* Worst Case denoted by big-o notation\n",
    "---\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "2o9hMlDButX7"
   },
   "source": [
    "**Creating a num list of range 100,000**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "num_list = [x for x in range(100000)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "id": "P1kPbp3FutX7"
   },
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   "source": [
    "import time\n",
    "\n",
    "#LINEAR SEARCH\n",
    "def LinearSearch(lys, element):\n",
    "    for i in range (len(lys)):\n",
    "        #sleep(1)\n",
    "        if lys[i] == element:\n",
    "            return i\n",
    "    return -1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "TCZKmOn1utX8"
   },
   "source": [
    "## <font color=blue> Best case scenario </font>"
   ]
  },
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   "cell_type": "code",
   "execution_count": 12,
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     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INPUT: 11\n",
      "Time taken:0.0s\n"
     ]
    }
   ],
   "source": [
    "num = int(input(\"INPUT: \"))\n",
    "start_time = time.time()\n",
    "res1 = LinearSearch(num_list, num)\n",
    "req_time_1 = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time_1,4)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "CD7OBUQputX9"
   },
   "source": [
    "## <font color=blue> Average case scenario </font>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "id": "MdFqP1vrutX-",
    "outputId": "89e8b44e-fd47-4ca8-e632-38802128692f"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INPUT: 500000\n",
      "Time taken:0.0038s\n"
     ]
    }
   ],
   "source": [
    "num = int(input(\"INPUT: \"))\n",
    "start_time = time.time()\n",
    "res2 = LinearSearch(num_list, num)\n",
    "req_time_2 = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time_2,4)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "7xweJfTyutX-"
   },
   "source": [
    "## <font color=blue> Worst case scenario </font>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "id": "tHHLX4WbutX-",
    "outputId": "1777588a-ca7e-475f-e042-32a3bac4f3f2"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "INPUT: 100000\n",
      "Time taken:0.004s\n"
     ]
    }
   ],
   "source": [
    "num = int(input(\"INPUT: \"))\n",
    "start_time = time.time()\n",
    "res3 = LinearSearch(num_list, num)\n",
    "req_time_3 = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time_3,4)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [],
   "source": [
    "def ReverseSearch(array,element):\n",
    "    n = len(array)\n",
    "    for i in range(-1,-n,-1):\n",
    "        print(array[i])\n",
    "        if array[i] == element:\n",
    "            return(n-i+1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "9\n",
      "8\n",
      "7\n",
      "6\n",
      "5\n",
      "4\n",
      "3\n",
      "2\n",
      "None\n"
     ]
    }
   ],
   "source": [
    "array = [1, 2, 3, 4, 5, 6, 7, 8, 9]\n",
    "element=1\n",
    "print(ReverseSearch(array,element))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "worst-case scenario: Search for 1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [],
   "source": [
    "def largestOddMultiple(number):\n",
    "    while(number%2==0):\n",
    "        sleep(1)\n",
    "        number = number/2\n",
    "        sleep(1)\n",
    "    return number"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time taken:0.0s\n"
     ]
    }
   ],
   "source": [
    "start_time = time.time()\n",
    "largestOddMultiple(133)\n",
    "req_time = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time,4)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time taken:2.0127s\n"
     ]
    }
   ],
   "source": [
    "start_time = time.time()\n",
    "largestOddMultiple(250)\n",
    "req_time = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time,4)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time taken:14.1295s\n"
     ]
    }
   ],
   "source": [
    "start_time = time.time()\n",
    "largestOddMultiple(128)\n",
    "req_time = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time,4)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time taken:0.0s\n"
     ]
    }
   ],
   "source": [
    "start_time = time.time()\n",
    "largestOddMultiple(175)\n",
    "req_time = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time,4)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time taken:0.0s\n"
     ]
    }
   ],
   "source": [
    "start_time = time.time()\n",
    "largestOddMultiple(133)\n",
    "req_time = time.time() - start_time\n",
    "\n",
    "print(\"Time taken:{}s\".format(round(req_time,4)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm Analysis"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def sum_n_numbers(n):\n",
    "    total = 0 # 1 assignment\n",
    "    for i in range(1,n+1): # n comparisons, n additions and n assignments\n",
    "        total = total+i # n additions and n assignments\n",
    "    return total"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "the above program is inefficient"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Important factors to consider:\n",
    "- Time: Execution time of a program\n",
    "- Space: Amount of space that a program occupies\n",
    "    \n",
    "In general, time is more important than space. Time complexity is more important than space complexity."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Asymptotic analysis is a mathematical concept performed to constrain the plot of our runtime under known expressions. In simple terms, we try to fit a curve to the runtime and express the runtime as a simple function of the input size. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Asymptotic analysis: \n",
    "- Evaluate algorithm performance in terms of input size\n",
    "- Calculate how time(or space) taken by algorithm grows with increase in input size"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Algorithm complexity\n",
    "- Specifying N number of operations that are performed in the algorithm\n",
    "- Big-O expresses expresses its runtime in terms of how quickly it expands relative to input size\n",
    "\n",
    "Algorithm complexity - It is defined as sum of all O[n] allocated to each line of code"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "NGnfDaDdTcUE"
   },
   "source": [
    "# **Linear Vs Binary**\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "4OCC-0uR_l4H"
   },
   "source": [
    "### <font color=blue> **Lets Compute the Time Complexity of Linear Search with Respect to Input Size**</font>"
   ]
  },
  {
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   "execution_count": 5,
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   "source": [
    "#Lets declare the global variable to save all records for visualization\n",
    "import pandas as pd  #importing Pandas \n",
    "\n",
    "# It will store {\"Size of Input\":[10,100,1000,10000], \"Search Time\":[0.1,0.2,0.3,0.4]}\n",
    "df = pd.DataFrame()"
   ]
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   "source": [
    "# Get Element by Linear Search\n",
    "import numpy as np\n",
    "temp = 10\n",
    "while temp<1000000:\n",
    "    ############### Create Array with Random Numbers #######\n",
    "    '''\n",
    "    #Set the size of the array\n",
    "    '''\n",
    "    temp = temp * 2\n",
    "    INPUT_SIZE = temp\n",
    "    arr = list(np.random.randint(low = 1,high=100000000,size=INPUT_SIZE))\n",
    "    arr = sorted(arr)\n",
    "    '''\n",
    "    #set the value that you want to search \n",
    "    [Example 1: for Best Case I want to search value of index 0]\n",
    "    [Example 2: for worst Case I want to search value of index \"Size of Array\"]\n",
    "    '''\n",
    "    val = arr[INPUT_SIZE-1]                                                    \n",
    "  ########################################################\n",
    "\n",
    "  ################ Linear Search Function ################\n",
    "    def LinearSearch(lys, element):\n",
    "        for i in range (len(lys)):\n",
    "            if lys[i] == element:\n",
    "                return i\n",
    "        return -1\n",
    "  ########################################################\n",
    "\n",
    "  ############### Measure Time ###########################\n",
    "    import time\n",
    "    start_time = time.time()\n",
    "    res = LinearSearch(arr, val)\n",
    "    req_time = time.time() - start_time\n",
    "  ########################################################\n",
    "\n",
    "  ###### Insert Time & Input size for Line graph ########\n",
    "\n",
    "    df = df.append({'Algorithm':\"Linear Search\",'InputSize': INPUT_SIZE, 'SearchTime':req_time}, ignore_index=True)\n",
    "\n",
    "############## Plot Line Graph #########################\n",
    "import plotly.express as px\n",
    "# print(df)\n",
    "# fig = px.line(x=RECORDS_LINEAR[\"InputSize\"], y=RECORDS_LINEAR[\"SearchTime\"], labels=dict(x=\"Input Size\", y=\"Serach Time\"))\n",
    "fig = px.line(df, x=\"InputSize\", y=\"SearchTime\", color = 'Algorithm', labels=dict(x=\"Input Size\", y=\"Search Time\"))\n",
    "\n",
    "fig.update_traces(mode='markers+lines')\n",
    "fig.update_layout(title_text='Linear Search Complexity O(n)', title_x=0.5)\n",
    "fig.show()\n",
    "########################################################"
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       "      <th>2</th>\n",
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       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>Linear Search</td>\n",
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       "      <td>Linear Search</td>\n",
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       "      <th>5</th>\n",
       "      <td>Linear Search</td>\n",
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       "      <td>0.000000</td>\n",
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       "    <tr>\n",
       "      <th>6</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>1280.0</td>\n",
       "      <td>0.000000</td>\n",
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       "    <tr>\n",
       "      <th>7</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>2560.0</td>\n",
       "      <td>0.000000</td>\n",
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       "    <tr>\n",
       "      <th>8</th>\n",
       "      <td>Linear Search</td>\n",
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       "      <td>0.000999</td>\n",
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       "      <th>9</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>10240.0</td>\n",
       "      <td>0.001000</td>\n",
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       "    <tr>\n",
       "      <th>10</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>20480.0</td>\n",
       "      <td>0.003000</td>\n",
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       "    <tr>\n",
       "      <th>11</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>40960.0</td>\n",
       "      <td>0.005996</td>\n",
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       "    <tr>\n",
       "      <th>12</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>81920.0</td>\n",
       "      <td>0.013446</td>\n",
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       "    <tr>\n",
       "      <th>13</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>163840.0</td>\n",
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       "    <tr>\n",
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       "      <td>327680.0</td>\n",
       "      <td>0.083442</td>\n",
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       "    <tr>\n",
       "      <th>15</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>655360.0</td>\n",
       "      <td>0.209793</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>16</th>\n",
       "      <td>Linear Search</td>\n",
       "      <td>1310720.0</td>\n",
       "      <td>0.333156</td>\n",
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       "        Algorithm  InputSize  SearchTime\n",
       "0   Linear Search       20.0    0.000000\n",
       "1   Linear Search       40.0    0.000000\n",
       "2   Linear Search       80.0    0.000000\n",
       "3   Linear Search      160.0    0.000000\n",
       "4   Linear Search      320.0    0.000000\n",
       "5   Linear Search      640.0    0.000000\n",
       "6   Linear Search     1280.0    0.000000\n",
       "7   Linear Search     2560.0    0.000000\n",
       "8   Linear Search     5120.0    0.000999\n",
       "9   Linear Search    10240.0    0.001000\n",
       "10  Linear Search    20480.0    0.003000\n",
       "11  Linear Search    40960.0    0.005996\n",
       "12  Linear Search    81920.0    0.013446\n",
       "13  Linear Search   163840.0    0.033362\n",
       "14  Linear Search   327680.0    0.083442\n",
       "15  Linear Search   655360.0    0.209793\n",
       "16  Linear Search  1310720.0    0.333156"
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   "source": [
    "### <font color=blue>**Now Lets compare the time complexity of Binary Search with Linear Search with same setup**</font>\n",
    "\n",
    "Below code will run multiple time with increasing size of the array to check how time complexity depends on inputs differently on different algorithms"
   ]
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      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Get Element by Linear Search\n",
    "import numpy as np\n",
    "\n",
    "temp = 10\n",
    "while temp<1000000:\n",
    "  ############### Create Array with Random Numbers #######\n",
    "    '''\n",
    "    #Set the size of the array\n",
    "    '''\n",
    "    temp = temp * 2\n",
    "    INPUT_SIZE = temp\n",
    "    arr = list(np.random.randint(low = 1,high=100000000,size=INPUT_SIZE))\n",
    "    arr = sorted(arr)\n",
    "    '''\n",
    "    #set the value that you want to search \n",
    "    [Example 1: for Best Case I want to search value of index 0]\n",
    "    [Example 2: for worst Case I want to search value of index \"Size of Array\"]\n",
    "    '''\n",
    "    val = arr[INPUT_SIZE-1]                                                    \n",
    "  ########################################################\n",
    "\n",
    "  ################ Binary Search Function ################\n",
    "    def BinarySearch(lys, val):\n",
    "        first = 0\n",
    "        last = len(lys)-1\n",
    "        index = -1\n",
    "        while (first <= last) and (index == -1):\n",
    "            mid = (first+last)//2\n",
    "            if lys[mid] == val:\n",
    "                index = mid\n",
    "            else:\n",
    "                if val<lys[mid]:\n",
    "                    last = mid -1\n",
    "                else:\n",
    "                    first = mid +1\n",
    "        return index\n",
    "  ########################################################\n",
    "\n",
    "\n",
    "\n",
    "  ############### Measure Time ###########################\n",
    "    import time\n",
    "    start_time = time.time()\n",
    "    res = BinarySearch(arr, val)\n",
    "    req_time = time.time() - start_time\n",
    "  ########################################################\n",
    "\n",
    "  ###### Insert Time & Input size for Line graph ########\n",
    "\n",
    "    df = df.append({'Algorithm':\"Binary Search\",'InputSize': INPUT_SIZE, 'SearchTime':req_time}, ignore_index=True)\n",
    "\n",
    "############## Plot Line Graph #########################\n",
    "import plotly.express as px\n",
    "# print(df)\n",
    "# fig = px.line(x=RECORDS_LINEAR[\"InputSize\"], y=RECORDS_LINEAR[\"SearchTime\"], labels=dict(x=\"Input Size\", y=\"Serach Time\"))\n",
    "fig = px.line(df, x=\"InputSize\", y=\"SearchTime\", color = 'Algorithm', labels=dict(x=\"Input Size\", y=\"Serach Time\"))\n",
    "\n",
    "fig.update_traces(mode='markers+lines')\n",
    "fig.update_layout(title_text='Binary Search Complexity O(log n)', title_x=0.5)\n",
    "fig.show()\n",
    "########################################################"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "the two curves of the plot clearly highlight how one algorithm is more efficient than the other. Asymptotic analysis reveals details that the simple approach fails to highlight."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Suppose you have a list of the first 100 natural numbers, which is 1 to 100. Now, a number is removed from the list at random. Given the list, you have to find the missing number. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Approach 1\n",
    "\n",
    "def findMissing(num_list):\n",
    "    for i in range(len(num_list)):\n",
    "        if num_list[i]!=i+1:\n",
    "            return i+1\n",
    "    return 100"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.0\n"
     ]
    }
   ],
   "source": [
    "import time\n",
    "num_list = np.arange(1,100)\n",
    "start_time = time.time()\n",
    "res = findMissing(num_list)\n",
    "req_time = time.time() - start_time\n",
    "print(req_time)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Approach 2\n",
    "\n",
    "\n",
    "def findMissing(num_list):\n",
    "    return(100*101/2 - sum(num_list))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.0\n"
     ]
    }
   ],
   "source": [
    "import time\n",
    "num_list = np.arange(1,100)\n",
    "start_time = time.time()\n",
    "res = findMissing(num_list)\n",
    "req_time = time.time() - start_time\n",
    "print(req_time)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This algorithm subtracts the sum of the list from the sum of the first 100 natural numbers. This would give us the missing number. This is more efficient"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "-uHXRJu_V9AD"
   },
   "source": [
    "# <font color=red> **Approach [Good vs Bad]** </font>\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "- List comprehension being faster than appending items to a list.\n",
    "- Strings can be concatenated using the join method instead of adding characters to an empty string.\n",
    "- The map function can be used to apply an operation over a list."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "xZq8SGNpVT2Y"
   },
   "source": [
    "# <font color=blue> **Use List Comprehension**\n",
    "**Because List comprehension is faster than append** </font>\n",
    "\n",
    "[**seconds** > **ms means miliseconds** > **us means microseconds** > **ns means nanoseconds**]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "-OpGbL6NcdT2"
   },
   "source": [
    "**Bad**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:09:47.677351Z",
     "start_time": "2021-09-16T02:09:42.588265Z"
    },
    "id": "U9ad9D-6RRpc",
    "outputId": "6e984a38-3c89-438e-b9c9-9bd88f098c28"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2.13 s ± 83.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "def odd(Number):\n",
    "    L = []\n",
    "    for i in range (1, Number+1):\n",
    "        if i%2 != 0:\n",
    "            L.append (i)\n",
    "    return L\n",
    "\n",
    "%timeit odd(9000000)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "W0tnle3Gcfkv"
   },
   "source": [
    "**Good**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:09:50.963490Z",
     "start_time": "2021-09-16T02:09:47.679351Z"
    },
    "id": "3cvb86JmRRpc",
    "outputId": "541ceba3-ad3e-46ab-b70e-4654dbc405da"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.7 s ± 109 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "def odd(Number):\n",
    "    L = [i for i in range (1, Number+1) if i%2 != 0]\n",
    "    return L\n",
    "\n",
    "%timeit odd(9000000)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "cjSTkdaLYW7A"
   },
   "source": [
    "### **Concatenate strings with join method**\n",
    "**Because + operators produce a new string and then copy the old information at each step, join() concatenates strings more quickly than +.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "_fQdkTbCciCA"
   },
   "source": [
    "**bad**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:09:57.958214Z",
     "start_time": "2021-09-16T02:09:50.966469Z"
    },
    "id": "AmbATw09Yrqg",
    "outputId": "f63c5424-5856-4e9a-aa21-e838334bb75f"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "5.19 ms ± 734 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)\n"
     ]
    }
   ],
   "source": [
    "def concat(lst):\n",
    "    concatenated = ''\n",
    "    for i in lst:\n",
    "        concatenated = concatenated + i\n",
    "\n",
    "lst = [\"A\"] * 10000\n",
    "\n",
    "%timeit concat(lst)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "QqKFAh57ckPx"
   },
   "source": [
    "**Good**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:07.444179Z",
     "start_time": "2021-09-16T02:09:57.960196Z"
    },
    "id": "liEL2-T2ZRiA",
    "outputId": "128f6652-b4f3-46df-c023-e551ea111629"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "136 µs ± 6.14 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n"
     ]
    }
   ],
   "source": [
    "def concat(lst):\n",
    "  concatenated = \"\".join(lst)\n",
    "\n",
    "lst = [\"A\"] * 10000\n",
    "\n",
    "%timeit concat(lst)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "TPRsi0ZManuy"
   },
   "source": [
    "### **Use Map Function**\n",
    "**Using built-in map function will make operations faster, because it's written in C.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "O4dYf-bAcnJK"
   },
   "source": [
    "**Bad**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:11.706738Z",
     "start_time": "2021-09-16T02:10:07.449177Z"
    },
    "id": "Pfd9nbTdarBJ",
    "outputId": "7892f792-0394-48c3-d16d-09d9633856fc"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "9.56 µs ± 417 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)\n"
     ]
    }
   ],
   "source": [
    "def square(numbers):\n",
    "    lst = []\n",
    "    for i in numbers:\n",
    "        lst.append(i**2)\n",
    "    \n",
    "numbers = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\n",
    "%timeit square(numbers)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "70fvyKgVcqYw"
   },
   "source": [
    "**Good**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.662049Z",
     "start_time": "2021-09-16T02:10:11.708737Z"
    },
    "id": "c9H-w7Lwbv0p",
    "outputId": "35a48434-991f-4c50-9ec3-a072be38e809"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "531 ns ± 55.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "numbers = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\n",
    "%timeit map(lambda x: x**2, numbers)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "narDrRfFocEO"
   },
   "source": [
    "# <font color=blue > **Time Complexity** </font>\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "yQw74k0PutYF"
   },
   "source": [
    "\n",
    "**How do you Calculate the Time Complexity of an Algorithm?**\n",
    "\n",
    "While comparing algorithms, we usually compare their worst case scenarios. \n",
    "Big O notation is used to define the time complexity of an algorithm\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Runtime of an algorithm depends on\n",
    "- Processor - Sinlge or multiprocessor\n",
    "- 32 or 64 bit architecture\n",
    "- Read and write speed of the machine\n",
    "- Time required to perform certain orpations(logical and arithmetic)\n",
    "- Return values and assignment\n",
    "- Input data"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Loops\n",
    "\n",
    "- Sequential - Add sequential statements\n",
    "- Nested loop - Multiply nested statements"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "```\n",
    "for i in range(n): # O(n)\n",
    "    for j in range(n): # O(n)\n",
    "        b[i][j] = a[i][j] # O(1)\n",
    "```\n",
    "        \n",
    "Time complexity = O(n) * O(n) * O(1) = O(n^2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### **Big O**\n",
    "* Big O notation is used to express an algorithm's upper bound in terms of Time Complexity.\n",
    "* As a result, Big O notation always reflects the algorithm's maximum duration for all input values.\n",
    "* As a result, Big O notation specifies the worst-case time complexity of an algorithm."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "WesR_HwCm5Hi"
   },
   "source": [
    "Note : n in the below cases, indicates the size of the input"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "C9RUbqThgXX8"
   },
   "source": [
    "# **Constant time – O(1)**\n",
    "\n",
    "\n",
    "**Example 1:**\n",
    "```\n",
    "print(x)                               // O(1)\n",
    "a = 10;                               // O(1)\n",
    "a = \"string\"                         // O(1)\n",
    "```\n",
    "**Example 2:**\n",
    "\n",
    "\n",
    "```\n",
    "a = input()                               // (1+1) = 2\n",
    "if a < 21 then start                      // 1\n",
    "      status = \"not allowed!\";            // 1\n",
    "end else start\n",
    "      status = \"Welcome!\";                // 1\n",
    "      visitor = visitor + 1;              // 1+1 = 2\n",
    "end;\n",
    "```\n",
    "**T(n) = 2 + 1 + max(1, 1+2) = 6 is the complexity of the given pseudo code. As a result, the big O remains constant T(n) = O(1).**\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "QpEHfAOhxkJT"
   },
   "source": [
    "### **Program to check number is even or odd**\n",
    "Find the time complexity of given code"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "id": "GSIDnb7XutYG"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "246 ns ± 31.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n",
      "100.0\n"
     ]
    }
   ],
   "source": [
    "# Get Element by Index\n",
    "def Even_Odd(n):\n",
    "    if n%2==0:\n",
    "        return \"Even\"\n",
    "    else:\n",
    "        return \"Odd\"\n",
    "\n",
    "    \n",
    "%timeit res = Even_Odd(10)\n",
    "print(res)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "273 ns ± 8.97 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n",
      "100.0\n"
     ]
    }
   ],
   "source": [
    "%timeit res = Even_Odd(10**10)\n",
    "print(res)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### operations are O(1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [],
   "source": [
    "list = [1,2,3,4,5,5,6,3,3,2,1,1,2,23,3]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "65.7 ns ± 5.5 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit len(list)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "315 ns ± 41.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit max(list)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "208 ns ± 12.6 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit sum(list)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "87.2 ns ± 8.96 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit list.append(0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "dCDYdxHWxM8K"
   },
   "source": [
    "#  **Linear time – O(n)**\n",
    "\n",
    "\n",
    "**Example 1:**\n",
    "\n",
    "\n",
    "```\n",
    "sum = 0;                                  // 1\n",
    "for i = 1 to n do begin                   // (1+1)*n = 2n\n",
    "      sum = sum + i;                      // (1+1)*n = 2n\n",
    "end;\n",
    "print(sum);                               // 1\n",
    "```\n",
    "**So, the complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).**\n",
    "\n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.708021Z",
     "start_time": "2021-09-16T02:10:14.695029Z"
    },
    "id": "YFCI-UMrwb9N"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "43.8 µs ± 1.55 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "100.0\n"
     ]
    }
   ],
   "source": [
    "def SumAll(N):\n",
    "    sum = 0\n",
    "    for i in range (1, N+1):\n",
    "        sum = sum + i\n",
    "    return sum\n",
    "\n",
    "val = 1000\n",
    "\n",
    "%timeit res = SumAll(val) \n",
    "print(res)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This is a rather inefficient method of finding the sum of first n natural numbers. The reason, it being O(n) is we have to iterate n times to find the sum.\n",
    "\n",
    " "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.739003Z",
     "start_time": "2021-09-16T02:10:14.714018Z"
    },
    "id": "2QBB1cL4Sa5m"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "221 ns ± 38.3 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)\n",
      "100.0\n"
     ]
    }
   ],
   "source": [
    "def SumAll(N):\n",
    "    return N * (N+1)//2\n",
    "\n",
    "val = 1000\n",
    "%timeit res = SumAll(val) \n",
    "print(res)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {},
   "outputs": [],
   "source": [
    "sentence = \"TAMAD ekfk dfk lmadl.\"\n",
    "def count_words_1(string):\n",
    "    for i in string:\n",
    "        count=1\n",
    "        if i=='':\n",
    "            count+=1\n",
    "        return count"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "183 ns ± 8.77 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit count_words_1(sentence)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {},
   "outputs": [],
   "source": [
    "def count_words_2(string):\n",
    "    for i in string:\n",
    "        count=1\n",
    "        if i=='':\n",
    "            count+=1\n",
    "        return count"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "189 ns ± 25.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%timeit count_words_2(sentence)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "UiOeSgAwxdfU"
   },
   "source": [
    "# **Quadratic time – O(n^2)**\n",
    "\n",
    "\n",
    "**Nested Loop (looping within looping): Because there is at least one looping inside the main looping, the statement's running time is used O(n^2) or O(n^3).**\n",
    "\n",
    "**Example 1**\n",
    "\n",
    "\n",
    "```\n",
    "for j in range(n):                          // (1+1)*n  = 2n\n",
    "   for k in range(n):                       // (1+1)n*n = 2n^2\n",
    "       x = x + 1;                           // (1+1)n*n = 2n^2\n",
    "       print(x);                            // (n*n)    = n^2\n",
    "   end;\n",
    "end;\n",
    "```\n",
    "\n",
    "**Example 2:**\n",
    "k = 3 means length of nested loop is 3\n",
    "\n",
    "\n",
    "```\n",
    "for i in range(n):\n",
    "  for j in range(n):\n",
    "    for k in range(n):\n",
    "      # some O(1) operation\n",
    "```\n",
    "the time complexity of the above code is O(n^k). Here k is 3. So the solution would be (n³).\n",
    "\n",
    "**Example 3**\n",
    "\n",
    "the following code consists of three phases with time complexities\n",
    "\n",
    "O ( n ), O ( n² ) and O ( n ). Thus, the total time complexity is O ( n² ).\n",
    "\n",
    "\n",
    "\n",
    "```\n",
    "  //Phase 1, time complexity is O(n)\n",
    "     for i in range(n):\n",
    "        #some code\n",
    "\n",
    "  //Phase 2 , time complexity is O(n²)\n",
    "     for j in range(n):\n",
    "          for k in range(n):\n",
    "              #some code\n",
    "\n",
    "  //Phase 3, time complexity is O(n)\n",
    "    for l in range(n):\n",
    "        #some code\n",
    "            \n",
    "```\n",
    "\n",
    "**How to combine time complexities of consecutive loops?**\n",
    "\n",
    "When there are multiple loops in a row, the time complexity is calculated as the sum of the time complexities of the individual loops.\n",
    "\n",
    "\n",
    "\n",
    "```\n",
    "   for i in range(m):\n",
    "        #some O(1) expressions\n",
    "   \n",
    "   for j in range(n):\n",
    "        #some O(1) expressions\n",
    "   \n",
    "   Time complexity of the code is O(m) + O(n) which is O(m+n)\n",
    "   If m == n, then time complexity will becomes O(2n) which is O(n).\n",
    "```\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "<img src=\"https://he-s3.s3.amazonaws.com/media/uploads/873c255.png\">"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "GrcuF5-8x6Uw"
   },
   "source": [
    "### <font color=green>  **Program to search element in 2D Array with linear search** </font>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.861934Z",
     "start_time": "2021-09-16T02:10:14.742002Z"
    },
    "id": "ZS0T02Q3xiYs"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.73 µs ± 120 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n",
      "(2, 3)\n"
     ]
    }
   ],
   "source": [
    "# Linear Search in 2D Array\n",
    "def MatrixLinearSearch(arr, rows, columns, search):\n",
    "  for i in range(rows):\n",
    "      for j in range(columns):\n",
    "        if arr[i][j]==search:\n",
    "          return (i, j)\n",
    "\n",
    "# Input\n",
    "arr = [[5,12,17,9,3],[13,4,8,14,1],[9,6,3,7,21]]\n",
    "col = 5\n",
    "rows = 3\n",
    "search = 7\n",
    "%timeit res = MatrixLinearSearch(arr, rows,col, search)\n",
    "print(res)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.77 s ± 236 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "n=1000\n",
    "\n",
    "import numpy as np\n",
    "search_element = n**2/2\n",
    "\n",
    "inp = np.arange(0,n**2).reshape((n,n))\n",
    "%timeit res = MatrixLinearSearch(inp, n,n, search_element)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "GCnnuYjQutYK"
   },
   "source": [
    "# **Other Orders**\n",
    "\n",
    "## <font color=blue> **$O(logn)$** </font>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "QvHHhmVvBg4N"
   },
   "source": [
    "O(log N) basically means time goes up linearly while the n goes up exponentially. So if it takes 1 second to \n",
    "compute 10 elements, it will take 2 seconds to compute 100 elements, 3 seconds to compute 1000 elements, and so on.\n",
    "\n",
    "It is O(log n) when we do divide and conquer type of algorithms e.g binary search. \n",
    "Another example is quick sort where each time we divide the array into two parts and \n",
    "each time it takes O(N) time to find a pivot element. Hence it N O(log N)\n",
    "\n",
    "Just in case somebody is missing it, in programming context, the base of log is 2 (not 10), so O(log n) scales like 1 sec for 10 elements, 2 sec for 20, 3 for 40 etc."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "id": "IP6EsFMUutYK",
    "outputId": "2e2f9e2b-9910-4c02-bce9-bb3eb7ea3a3d",
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "118 ns ± 25.6 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "378 ns ± 40.4 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "421 ns ± 56.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "462 ns ± 45.9 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "570 ns ± 65.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "711 ns ± 91.9 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "678 ns ± 88.7 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "766 ns ± 73.1 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "810 ns ± 181 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "835 ns ± 51.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "900 ns ± 84.4 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.09 µs ± 89.9 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.1 µs ± 163 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.16 µs ± 99.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.4 µs ± 182 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.42 µs ± 171 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.44 µs ± 199 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.7 µs ± 264 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.81 µs ± 336 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.73 µs ± 241 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.89 µs ± 158 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "2.39 µs ± 283 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "2.65 µs ± 517 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "2.68 µs ± 320 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "3.29 µs ± 301 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "3.47 µs ± 417 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "3.72 µs ± 511 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "3.87 µs ± 379 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "4.74 µs ± 221 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "4.47 µs ± 369 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "import time\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "# the order is log(n) because the value of i increases geometrically\n",
    "def loop(n):\n",
    "    i = 1\n",
    "    while i<n:\n",
    "        i = i*10\n",
    "        \n",
    "run_times_lower = []\n",
    "run_times_upper = []\n",
    "num = []\n",
    "for i in range(30):\n",
    "    num.append(i)\n",
    "    time = %timeit -o -n 10000 loop(10**i)\n",
    "    run_times_lower.append(time.average - time.stdev)\n",
    "    run_times_upper.append(time.average + time.stdev)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "id": "sBFI_qubutYL",
    "outputId": "1517ca71-d7dd-47de-e55e-2607f9ec7c18"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "<matplotlib.collections.PolyCollection at 0x2062829ad30>"
      ]
     },
     "execution_count": 47,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(run_times_lower)\n",
    "plt.plot(run_times_upper)\n",
    "plt.fill_between(num,run_times_lower, run_times_upper, color='pink',\n",
    "                 alpha=0.2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "XOc1c7NPutYL"
   },
   "source": [
    "## <font color=blue> $O(2^n)$ </font>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {
    "id": "Jf7ahzIKutYL",
    "outputId": "b44914e4-5e15-4d95-88c6-c93f9de0c595"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "126 ns ± 22.1 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "391 ns ± 34.5 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "1.14 µs ± 203 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "2.4 µs ± 293 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "3.75 µs ± 274 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "6.39 µs ± 635 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "10.7 µs ± 1.22 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "17.8 µs ± 758 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n",
      "33.3 µs ± 3.78 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n"
     ]
    }
   ],
   "source": [
    "#Use recursion\n",
    "def fibonacci(n):\n",
    "    if n==0 or n==1:\n",
    "        return 1\n",
    "    return(fibonacci(n-1)+fibonacci(n-2))\n",
    "\n",
    "run_times_lower = []\n",
    "run_times_upper = []\n",
    "num = []\n",
    "for i in range (1,10):\n",
    "    num.append(i)\n",
    "    time = %timeit -o -n 10000 fibonacci(i)\n",
    "    run_times_lower.append(time.average - time.stdev)\n",
    "    run_times_upper.append(time.average + time.stdev)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {
    "id": "o_nCfOkQutYM",
    "outputId": "7fee0c59-97e7-4bdd-d4d8-3a9bc44ed62e"
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "<matplotlib.collections.PolyCollection at 0x20628a3c1c0>"
      ]
     },
     "execution_count": 49,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "import math\n",
    "import time\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "plt.plot(num,run_times_lower)\n",
    "plt.plot(num,run_times_upper)\n",
    "plt.fill_between(num,run_times_lower, run_times_upper, color='pink',\n",
    "                 alpha=0.2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Compare O(1) , O(n) and O(n^2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {
    "id": "KzYU2oyBBg4T"
   },
   "outputs": [],
   "source": [
    "def get_value(array):\n",
    "    return array[-1]\n",
    "\n",
    "def search_value(array):\n",
    "    for index, value in enumerate(array):\n",
    "      if index == len(array)-1:\n",
    "        return index\n",
    "\n",
    "def MatrixLinearSearch(arr):\n",
    "  arr = np.reshape(array, (-1, len(arr)))\n",
    "  rows, columns = arr.shape\n",
    "  for i in range(rows):\n",
    "      for j in range(columns):\n",
    "        if arr[i][j]==500:\n",
    "          return (i, j)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {
    "id": "14G2S_fFBg4X",
    "outputId": "6a04b0e6-e5d6-4c34-8f84-8da81967ccbd"
   },
   "outputs": [],
   "source": [
    "import pandas as pd\n",
    "import time\n",
    "from numpy.random import randint\n",
    "import numpy as np\n",
    "\n",
    "\n",
    "DATA = pd.DataFrame()\n",
    "\n",
    "for array_size in range(1, 100000, 10000):\n",
    "  array = randint(0, 10, array_size)\n",
    "\n",
    "  # Big O(1)\n",
    "  start = time.time()\n",
    "  get_value(array)\n",
    "  end = time.time()\n",
    "  req_time = end - start\n",
    "  DATA = DATA.append({'Time Complexity':\"O(1)\",'InputSize': array_size, 'Time':req_time}, ignore_index=True)\n",
    "\n",
    "\n",
    "  # Big O(n)\n",
    "  start = time.time()\n",
    "  search_value(array)\n",
    "  end = time.time()\n",
    "  req_time = end - start\n",
    "  DATA = DATA.append({'Time Complexity':\"O(n)\",'InputSize': array_size, 'Time':req_time}, ignore_index=True)\n",
    "\n",
    "\n",
    "  # Big O(n^2)\n",
    "  start = time.time()\n",
    "  MatrixLinearSearch(array)\n",
    "  end = time.time()\n",
    "  req_time = end - start\n",
    "  DATA = DATA.append({'Time Complexity':\"O(n^2)\",'InputSize': array_size, 'Time':req_time}, ignore_index=True)\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "id": "r8MyGt2rBg4Y"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "   Time Complexity  InputSize      Time\n",
      "0             O(1)        1.0  0.000000\n",
      "1             O(n)        1.0  0.001007\n",
      "2           O(n^2)        1.0  0.001389\n",
      "3             O(1)    10001.0  0.000000\n",
      "4             O(n)    10001.0  0.006593\n",
      "5           O(n^2)    10001.0  0.012159\n",
      "6             O(1)    20001.0  0.000000\n",
      "7             O(n)    20001.0  0.008997\n",
      "8           O(n^2)    20001.0  0.015489\n",
      "9             O(1)    30001.0  0.000000\n",
      "10            O(n)    30001.0  0.014004\n",
      "11          O(n^2)    30001.0  0.031405\n",
      "12            O(1)    40001.0  0.000000\n",
      "13            O(n)    40001.0  0.019712\n",
      "14          O(n^2)    40001.0  0.031324\n",
      "15            O(1)    50001.0  0.000000\n",
      "16            O(n)    50001.0  0.022548\n",
      "17          O(n^2)    50001.0  0.041743\n",
      "18            O(1)    60001.0  0.000000\n",
      "19            O(n)    60001.0  0.026731\n",
      "20          O(n^2)    60001.0  0.049785\n",
      "21            O(1)    70001.0  0.000000\n",
      "22            O(n)    70001.0  0.034571\n",
      "23          O(n^2)    70001.0  0.061106\n",
      "24            O(1)    80001.0  0.000000\n",
      "25            O(n)    80001.0  0.037272\n",
      "26          O(n^2)    80001.0  0.075157\n",
      "27            O(1)    90001.0  0.000000\n",
      "28            O(n)    90001.0  0.040552\n",
      "29          O(n^2)    90001.0  0.091884\n"
     ]
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      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "print(DATA)\n",
    "############## Plot Line Graph #########################\n",
    "import plotly.express as px\n",
    "# print(df)\n",
    "# fig = px.line(x=RECORDS_LINEAR[\"InputSize\"], y=RECORDS_LINEAR[\"SearchTime\"], labels=dict(x=\"Input Size\", y=\"Serach Time\"))\n",
    "fig = px.line(DATA, x=\"InputSize\", y=\"Time\", color = 'Time Complexity', labels=dict(x=\"Input Size\", y=\"Time\"))\n",
    "\n",
    "fig.update_traces(mode='markers+lines')\n",
    "fig.update_layout(title_text='Time Complexity', title_x=0.5)\n",
    "fig.show()\n",
    "# fig.show(renderer='colab')\n",
    "########################################################"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "bjt2gVMBzgnY"
   },
   "source": [
    "### **In the below table you can see computation required for different input size**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "lsyXNIgmrXyU"
   },
   "source": [
    "<img src=\"https://dz2cdn1.dzone.com/storage/temp/5068673-screen-shot-2017-04-24-at-21022-pm.png\">"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "P_iS308LcP9F"
   },
   "source": [
    "# <font color=blue> **Space Complexity** </font>\n",
    "---\n",
    "<font color=\"red\">Consider a basic python procedure that adds two integers to get you started (numbers without a fractional part):</font>\n",
    "\n",
    "\n",
    "```\n",
    "def sum(a,b):\n",
    "  return a+b\n",
    "\n",
    "#In this particular method, three variables are used and allocated in memory:\n",
    "\n",
    "#the first int argument, a\n",
    "#the second int argument, b\n",
    "#and the returned sum result which is also an int like a and b\n",
    "```\n",
    "In this example, we have three integer variables. this procedure requires 12 bytes of memory (3*4 bytes).\n",
    "\n",
    "We can see that the space complexity is constant, hence it can be written as **O(1) in big-O notation**.\n",
    "\n",
    "\n",
    "---\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "<font color=\"red\">Now, let’s determine the space complexity of an program that sums all integer in an array:</font>\n",
    "\n",
    "\n",
    "```\n",
    "def sum_of(array):\n",
    "  n = len(array)\n",
    "  sum = 0\n",
    "\n",
    "  for i in range(n):\n",
    "    sum = sum + array[i]\n",
    "\n",
    "  return sum\n",
    "```\n",
    "**Explanation:**\n",
    "\n",
    "In the above-given code, the array consists n integer. So, the space required by the array is 4 * n. \n",
    "\n",
    "There are also integer variables like n, i and sum. The total space occupied by the programme is 4n + 12 bytes, assuming each variable is 4 bytes. The space complexity is n because the largest order of n in the equation 4n + 12 is **O(n) or linear**..\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "---\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "XbOeJ_F59cR9"
   },
   "source": [
    "## **Good vs Bad Approach For Space Complexities**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "4QLEK_ys9qQR"
   },
   "source": [
    "**Looping over**\n",
    "\n",
    "range is better than [list, set, dict, str] as a iterable, because range function take constant space for any range as shown below."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "UQK4CZ2-CBff"
   },
   "source": [
    "**Bad**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.892917Z",
     "start_time": "2021-09-16T02:10:14.879923Z"
    },
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "4zUaVKcRCKlA",
    "outputId": "cdc614da-154b-47ba-f064-b998092f1573"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "58\n",
      "112\n",
      "152\n",
      "728\n",
      "232\n"
     ]
    }
   ],
   "source": [
    "import sys\n",
    "   \n",
    "b=sys.getsizeof('123456789')\n",
    "print(b)\n",
    "  \n",
    "c=sys.getsizeof((1,2,3,4,5,6,7,8,9))\n",
    "print(c)\n",
    "   \n",
    "d=sys.getsizeof([1,2,3,4,5,6,7,8,9])\n",
    "print(d)\n",
    "   \n",
    "e=sys.getsizeof({1,2,3,4,5,6,7,8,9})\n",
    "print(e)\n",
    "   \n",
    "f=sys.getsizeof({1:'a',2:'b',3:'c',4:'d'})\n",
    "print(f)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "ILMM_FE2Eg8W"
   },
   "source": [
    "**Good**\n",
    "\n",
    "using range we can create iterable with constant space complexity"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.922898Z",
     "start_time": "2021-09-16T02:10:14.895914Z"
    },
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "ADL55U2u9lfb",
    "outputId": "34770afc-6038-4e08-b0b9-9b0e490bea40"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "48\n"
     ]
    }
   ],
   "source": [
    "# To iterate over 10 elements it will take \n",
    "import sys\n",
    "a=sys.getsizeof(range(10)) \n",
    "print(a)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.938890Z",
     "start_time": "2021-09-16T02:10:14.925897Z"
    },
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "KIiaTZuPABXQ",
    "outputId": "2edbbb31-eb0a-469b-b459-cce2ba39f491"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "48\n"
     ]
    }
   ],
   "source": [
    "# To iterate over 100000 elements it will take same space\n",
    "import sys\n",
    "a=sys.getsizeof(range(100000)) \n",
    "print(a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "An important observation to be made is that dict occupies the most amount of space. However, this is to be expected, given that dict is the most time-efficient data structure of all the other data structures for operations, such as searching, sorting, adding and deleting.\n",
    "\n",
    " "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Mtv2IurdhY8G"
   },
   "source": [
    "# **Summary**\n",
    "Finally, we discussed the spatial and temporal complexity considerations. We now know that maximising the algorithm's space and time complexity is impossible. Those variables should be adjusted based on our needs and the surroundings."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Mathematical induction"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. Prove that a statement holds true for n=1\n",
    "2.  Prove that a statement holds true for n=k\n",
    "3.  Prove that a statement holds true for n=k+1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "- Prove that P(1) is true.\n",
    "- Assume that P(n) is true for n = k-1.\n",
    "- Prove that P(n) is true for n = k."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here, k is a natural number. P(1) indicates the proof, that is, whether the left-hand side expression equates to that on the right-hand side. \n",
    "The logic behind the proof of mathematical induction is that we know that if P(k-1) is true, then P(k) is true. Since k can be any natural number, this can be extended for P(k+1), P(k+2) and so on.\n",
    "So, to prove that P(n) is true for any n, we need to prove that it is true for one natural number, which we call the 'Base case'.\n",
    "\n",
    " \n",
    "\n",
    "We usually choose this base case to be for n=1."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mathematical problems are solved by induction and implemented by recursion."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "gWEImrNOTKd5"
   },
   "source": [
    "**Proof By Induction:**\n",
    "\n",
    "Sum of first N odd numbers\n",
    "> ∑(2i – 1) = n^2 where i varies from 1 to n\n",
    "\n",
    "\n",
    "<font color=\"red\">**We may use the odd number theorem to get the sum of the first n odd numbers, which asserts that the sum of the first n odd numbers equals the square of n.**</font>\n",
    "\n",
    "**Example 1**:\n",
    "\n",
    "* Basis: The proposition P(1) is true.\n",
    "\n",
    "* Inductive Step:\n",
    "P(n) --> P(n+1), is true for all p\n",
    "\n",
    "If n equals 10, the sum of the first ten odd numbers is 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100.\n",
    "If the odd number theorem is applied, the sum of the first ten odd numbers equals n * n = 10 * 10 = 100.\n",
    "The following is how the above strategy is put into action: \n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "x-nE_kaYGYuP"
   },
   "source": [
    "\n",
    "**How does it work?**\n",
    "\n",
    "We can prove it using mathematical induction. We know it is true for n = 1 and n = 2 as sums are 1 and 4 (1 + 3) respectively.\n",
    "\n",
    "\n",
    "```\n",
    "Let it be true for n = k-1.\n",
    "\n",
    "Sum of first k odd numbers = \n",
    "  Sumation of first k-1 odd numbers + k'th odd number\n",
    "= (k-1)^2 + (2k - 1)\n",
    "= (k^2 - 2k + 1) + (2k - 1)\n",
    "= k^2\n",
    "```\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Q3lrz1T0G-aC"
   },
   "source": [
    "**Example 2**: \n",
    "\n",
    "**Proof by induction: Show that 9^n−2^n for any n natural number is divisible by 7.**\n",
    "```\n",
    "Let f(k)=9^k−2^k where k∈N.\n",
    "\n",
    "Base Case: f(1)=9−2=7=7(1).\n",
    "\n",
    "Inductive Hypothesis: Assume f(k) is divisible by 7 for some natural number k.\n",
    "\n",
    "f(k+1)=9^k+1−2^k+1=9⋅9^k−2⋅2^k.\n",
    "\n",
    "Inductive step: f(k+1)−2f(k)=7⋅9^k=7⋅9^k. So f(k+1)=7⋅9^k+2f(k).\n",
    "\n",
    "Recall: f(k) was assumed to be divisible by 7 for the inductive hypothesis. So we are done since f(k+1) is the sum of two things divisible by 7.\n",
    "```\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "ZrrMZcb117ju"
   },
   "source": [
    "# <font color=red>  **Recursion** </font>\n",
    "\n",
    "### The connection between recursion and induction\n",
    "Eariler, you saw how to solve the problem of sum of the first 'n' odd natural numbers = n^2. Now let us look at how to calculate the sum of first 'n' odd natural numbers."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:10:14.954879Z",
     "start_time": "2021-09-16T02:10:14.942887Z"
    },
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "B9Sy3jldGNeS",
    "outputId": "9d0eef2b-ea25-4380-9f70-0d54867da192"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Sum of first 10 Odd Numbers is:  100\n"
     ]
    }
   ],
   "source": [
    "# Python3 program to find sum\n",
    "# of first n odd numbers\n",
    "\n",
    "def oddSum(n) :\n",
    "    \n",
    "    # Base case =>We know oddSum(1) = 1\n",
    "    if n==1:\n",
    "        return 1\n",
    "    \n",
    "    # Recursive step\n",
    "    # nth odd number = 2n-1\n",
    "    # We know oddSum(n) = oddSum(n-1)+ nth odd number\n",
    "    return (2*n-1)+oddSum(n-1)\n",
    "\n",
    "\n",
    "n = 10\n",
    "print (\" Sum of first\" , n, \"Odd Numbers is: \",oddSum(n) )"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "There are two types of recursions, which are as follows:\n",
    "\n",
    "- Direct recursion: Function A() calls itself.\n",
    "- Indirect recursion: Function A() calls Function B(), which in return calls Function A()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "120"
      ]
     },
     "execution_count": 57,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# factorial\n",
    "def factorial(n):\n",
    "    if n==1:\n",
    "        return n\n",
    "    return n*factorial(n-1)\n",
    "\n",
    "factorial(5)\n",
    "\n",
    "# factorial(5) = 5 *factorial(4)\n",
    "# factorial(5) = 5 * 4 *factorial(3)\n",
    "# factorial(5) = 5 * 4 * 3 *factorial(2)\n",
    "# factorial(5) = 5 * 4 * 3 * 2* factorial(1)\n",
    "# factorial(5) = 5 * 4 * 3 * 2* 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "fJSbxIxiGW2n"
   },
   "source": [
    "## **Use of Recursion**\n",
    "Recursion works best with what I like to call **\"fractal problems\"**, where you're dealing with a big thing that's made of smaller versions of that big thing, each of which is an even smaller version of the big thing, and so on. If you ever have to traverse or search through something like a tree or nested identical structures, you've got a problem that might be a good candidate for recursion.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "jKoG3C0DNCyf"
   },
   "source": [
    "To make our task easier and more practical, we break down a problem into smaller subproblems.\n",
    "\n",
    "<font color=\"blue\">The application of recursion is vital in computer science</font>\n",
    "\n",
    "* This is the foundation of AI. The NP problem cannot be solved in general, although it can be addressed to a certain extent (not totally) using recursion by limiting the depth of recursion.\n",
    "* Without recursion, the most important data structure, the Tree, does not exist. We can solve this in an iterative manner as well, but it will be a difficult effort.\n",
    "* Recursion is used in many well-known sorting algorithms (Quick sort, Merge sort, and so on).\n",
    "* Recursion is used extensively in all puzzle games (Chess, Candy Crush, and so on). Recursion has a plethora of applications these days since it is the backbone of searching, which is the most important thing.\n",
    "\n",
    "\n",
    "Recursion is a special form of iteration **in programming**, in which the call stack is used as a special mechanism of storing state. Any recursive algorithm can be rewritten to be iterative. It may be more difficult or less elegant, but the result is the same.\n",
    "\n",
    "There are various problems **in mathematics** that require recursive strategies to solve - for example, finding roots (Newton's Method), computing primes, graph optimization, and so on. However, even here, the distinction between the terms \"iteration\" and \"recursion\" is merely a matter of semantics."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "oUidcVDeSaVw"
   },
   "source": [
    "## <font color=\"red\">Fibonacci Series</font>\n",
    "Fibonacci series = 1,1,2,3,5,8,13...\n",
    "\n",
    "It is defined by\n",
    "- f(n) = f(n-1) + f(n-2) for n>2 where n is a natural number\n",
    "- f(2) = 1\n",
    "- f(1) = 1\n",
    "\n",
    "\n",
    "## <font color=\"red\">Fibonacci Series using Recursion</font>\n",
    "\n",
    "For a given value of 'n' print the first 'n' fibonacci number using recursion\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:12:08.514201Z",
     "start_time": "2021-09-16T02:10:14.956878Z"
    },
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "5EMg7rDwT_hG",
    "outputId": "3b51ceb6-2578-470b-a963-451d122205a6"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Enter a number, N, N>=2 : 100\n",
      "318 ns ± 12.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)\n"
     ]
    }
   ],
   "source": [
    "fibo_series = [0, 1]\n",
    "\n",
    "def fibonacci(n):\n",
    "    if n<=len(fibo_series) and n>0:\n",
    "        return fibo_series[n-1]\n",
    "    else:\n",
    "        fn = fibonacci(n-1) + fibonacci(n-2)\n",
    "        if n>len(fibo_series):\n",
    "            fibo_series.append(fn)\n",
    "        return fn\n",
    "\n",
    "n = int(input('Enter a number, N, N>=2 : '))\n",
    "\n",
    "# import time\n",
    "# start_time = time.time()\n",
    "%timeit fibonacci(n)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Slun7n2FWPhC"
   },
   "source": [
    "## <font color=\"red\">Fibonacci Series using Loop</font>\n",
    "For a given value of 'n' print the first 'n' fibonacci number using iteration\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 60,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:12:08.520200Z",
     "start_time": "2021-09-16T02:12:08.520200Z"
    },
    "id": "1XAr6sZ_VK79"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Enter a number, N, N>=2 : 100\n",
      "18.1 µs ± 1.11 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)\n"
     ]
    }
   ],
   "source": [
    "n = int(input('Enter a number, N, N>=2 : '))\n",
    "\n",
    "def fib(n):\n",
    "  num = 0\n",
    "  num1 = 0\n",
    "  num2 = 1\n",
    "  count = 0\n",
    "  result = []\n",
    "  while (count<n):\n",
    "      result.append(num1)\n",
    "      num = num1 +num2\n",
    "      num1 = num2\n",
    "      num2 = num\n",
    "      count +=1\n",
    "  return result\n",
    "\n",
    "\n",
    "# import time\n",
    "# start_time = time.time()\n",
    "%timeit result = fib(n)\n",
    "# req_time = time.time() - start_time\n",
    "\n",
    "# print(\"Time:\",req_time, \"\\n\", result)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Recursive solution is faster"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {
    "id": "xmQE91y1Bg4v"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The gcd of 60 and 48 is : 12000\n",
      "Wall time: 0 ns\n"
     ]
    }
   ],
   "source": [
    "%%time\n",
    "def GCD(a,b):\n",
    "    if(b==0):\n",
    "        return a\n",
    "    else:\n",
    "        return GCD(b,a%b)\n",
    "\n",
    "a = 60000\n",
    "b = 48000\n",
    "\n",
    "print (\"The gcd of 60 and 48 is : \",end=\"\")\n",
    "print (GCD(a,b))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {
    "id": "pckq-wPqBg4w"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The gcd of 60 and 48 is : 12000\n",
      "Wall time: 500 µs\n"
     ]
    }
   ],
   "source": [
    "%%time\n",
    "def GCD1(x, y):\n",
    "\n",
    "  while(y):\n",
    "    x, y = y, x % y\n",
    "\n",
    "  return x\n",
    "\n",
    "a1 = 60000\n",
    "b1 = 48000\n",
    "\n",
    "# prints 12\n",
    "print (\"The gcd of 60 and 48 is : \",end=\"\")\n",
    "print (GCD(a1,b1))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "FwqJFuChBg4w"
   },
   "source": [
    "## <font color=red> Plot iteration vs recursion </font>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 65,
   "metadata": {
    "id": "iVCjuiBbBg4w"
   },
   "outputs": [],
   "source": [
    "import pandas as pd\n",
    "import time\n",
    "\n",
    "\n",
    "series = pd.DataFrame()\n",
    "\n",
    "# Loop\n",
    "def fib1(n):\n",
    "    a = 0\n",
    "    b = 1\n",
    "    if n == 0:\n",
    "        return a\n",
    "    elif n == 1:\n",
    "        return b\n",
    "    else:\n",
    "        for i in range(2, n):\n",
    "            c = a + b\n",
    "            a = b\n",
    "            b = c\n",
    "        return b\n",
    "\n",
    "\n",
    "# Recursion\n",
    "def fib(n):\n",
    "    Array = [0, 1]\n",
    "    if n<= len(Array):\n",
    "        return Array[n-1]\n",
    "    else:\n",
    "        temp_fib = fib(n-1)+fib(n-2)\n",
    "        Array.append(temp_fib)\n",
    "        return temp_fib\n",
    "\n",
    "\n",
    "for a in range(10, 40, 4):\n",
    "\n",
    "  start = time.time()\n",
    "  ans = fib(a)\n",
    "  end = time.time()\n",
    "  req_time = end - start\n",
    "  series = series.append({'Algorithm':\"Recursion\",'InputSize': a, 'Time':req_time, \"Answer\":ans}, ignore_index=True)\n",
    "\n",
    "\n",
    "\n",
    "  start = time.time()\n",
    "  ans = fib1(a)\n",
    "  end = time.time()\n",
    "  req_time = end - start\n",
    "  series = series.append({'Algorithm':\"Loop\",'InputSize': a, 'Time':req_time, \"Answer\":ans}, ignore_index=True)\n",
    "\n"
   ]
  },
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   "execution_count": 66,
   "metadata": {
    "id": "kbFdjXaqBg4x"
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      "    Algorithm  InputSize       Time      Answer\n",
      "0   Recursion       10.0   0.000000        34.0\n",
      "1        Loop       10.0   0.000000        34.0\n",
      "2   Recursion       14.0   0.000000       233.0\n",
      "3        Loop       14.0   0.000000       233.0\n",
      "4   Recursion       18.0   0.002000      1597.0\n",
      "5        Loop       18.0   0.000000      1597.0\n",
      "6   Recursion       22.0   0.010849     10946.0\n",
      "7        Loop       22.0   0.000000     10946.0\n",
      "8   Recursion       26.0   0.074584     75025.0\n",
      "9        Loop       26.0   0.000000     75025.0\n",
      "10  Recursion       30.0   0.580668    514229.0\n",
      "11       Loop       30.0   0.000000    514229.0\n",
      "12  Recursion       34.0   6.745952   3524578.0\n",
      "13       Loop       34.0   0.000000   3524578.0\n",
      "14  Recursion       38.0  30.084532  24157817.0\n",
      "15       Loop       38.0   0.000000  24157817.0\n"
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       "                            \n",
       "var gd = document.getElementById('c16aad40-9abd-4dab-ae58-5d4de8470443');\n",
       "var x = new MutationObserver(function (mutations, observer) {{\n",
       "        var display = window.getComputedStyle(gd).display;\n",
       "        if (!display || display === 'none') {{\n",
       "            console.log([gd, 'removed!']);\n",
       "            Plotly.purge(gd);\n",
       "            observer.disconnect();\n",
       "        }}\n",
       "}});\n",
       "\n",
       "// Listen for the removal of the full notebook cells\n",
       "var notebookContainer = gd.closest('#notebook-container');\n",
       "if (notebookContainer) {{\n",
       "    x.observe(notebookContainer, {childList: true});\n",
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       "    x.observe(outputEl, {childList: true});\n",
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      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "print(series)\n",
    "############## Plot Line Graph #########################\n",
    "import plotly.express as px\n",
    "# print(df)\n",
    "# fig = px.line(x=RECORDS_LINEAR[\"InputSize\"], y=RECORDS_LINEAR[\"SearchTime\"], labels=dict(x=\"Input Size\", y=\"Serach Time\"))\n",
    "fig = px.line(series, x=\"InputSize\", y=\"Time\", color = 'Algorithm', labels=dict(x=\"Input Size\", y=\"Time\"))\n",
    "\n",
    "fig.update_traces(mode='markers+lines')\n",
    "fig.update_layout(title_text='Recursion vs Loops', title_x=0.5)\n",
    "fig.show()\n",
    "# fig.show(renderer='colab')\n",
    "########################################################"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "p53MAc0rW-oh"
   },
   "source": [
    "<img src=\"http://2.bp.blogspot.com/-AxaPtxd0Z1c/Usn2a3gTGyI/AAAAAAAAA_c/BIbMy6nn2ZE/s1600/plot.png\">"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Iteration more efficient in terms of time complexity/runtime"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 77,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5"
      ]
     },
     "execution_count": 77,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Returns the value of floor(log2(number))\n",
    "\n",
    "def Unknown_func(number):\n",
    "    if(number == 1):\n",
    "        return (0)\n",
    "    else:\n",
    "        return (1 + Unknown_func(number//2))\n",
    "    \n",
    "Unknown_func(63)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 70,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5"
      ]
     },
     "execution_count": 70,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "11//2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Given a positive integer 'n' and another positive integer 'k' (<n),  print the following pattern using recursion.\n",
    "\n",
    "# Input 1: n = 12, k = 5\n",
    "# Output 2: 12, 7, 2, -3, 2, 7, 12\n",
    "\n",
    "\n",
    "# Input 2: n = 10, k =2\n",
    "# Output 2: 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10\n",
    "\n",
    "# Basically, you need to keep decrementing the given value of 'n' by 'k' until you encounter 0 or a negative number, in which case, you need to start incrementing by 'k' until you reach 'n' again."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "n=12\n",
    "k=5\n",
    "m=n\n",
    "def pattern(n,m,k):\n",
    "    if m<=0:\n",
    "        print(m, end=\", \")\n",
    "        return\n",
    "    \n",
    "    print(m, end = \", \")\n",
    "    pattern(n,m-k,k)\n",
    "    \n",
    "    if(n != m):\n",
    "        print(m, end = \", \")\n",
    "    else:\n",
    "        print(m)\n",
    "        \n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "12, 7, 2, -3, 2, 7, 12\n"
     ]
    }
   ],
   "source": [
    "pattern(n, m, k)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10"
     ]
    }
   ],
   "source": [
    "n=10\n",
    "k=2\n",
    "\n",
    "num=n\n",
    "\n",
    "def decrement(n,k):  \n",
    "    if n<=0:\n",
    "        return n\n",
    "    print(n, end=\", \")\n",
    "    return decrement(n-k,k)\n",
    "    \n",
    "def increment(n,k):\n",
    "    \n",
    "    if n!=num:\n",
    "        print(n, end=\", \")\n",
    "        return increment(n+k,k)\n",
    "    else:\n",
    "        print(n, end=\"\")\n",
    "        return\n",
    "\n",
    "def pattern(n,k):\n",
    "    n=decrement(n,k)\n",
    "    increment(n,k)\n",
    "\n",
    "pattern(n,k)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "12\n",
      "5\n",
      "12, 7, 2, -3, 2, 7, 12\n"
     ]
    }
   ],
   "source": [
    "# Reading the input\n",
    "n = int(input())\n",
    "k = int(input())\n",
    "\n",
    "# Define the function\n",
    "def pattern(n, m, k):\n",
    "    \n",
    "    # Base case. If m <= 0, print m and return from the function\n",
    "    if m <= 0:\n",
    "        print(m, end = \", \")\n",
    "        return\n",
    "    \n",
    "    # Print m, call the recursive function with (n, m-k, k), and then print m again\n",
    "    print(m, end = \", \")\n",
    "    pattern(n, m-k, k)\n",
    "    # This conditional statement has been used so that you do not print a comma \n",
    "    # after the last element.\n",
    "    if(n != m):\n",
    "        print(m, end = \", \")\n",
    "    else:\n",
    "        print(m)\n",
    "\n",
    "# Declaring another variable m equal to n\n",
    "m = n\n",
    "\n",
    "# Calling the function with 3 arguments - n, m, and k. The extra m will be used\n",
    "# for recursive calls. The n will be preserved because if you check the output\n",
    "# format, you will see that after the last element, you shouldn't print a comma.\n",
    "# The value of 'n' in the function will be used primarily for that.\n",
    "pattern(n, m, k)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "zh7388NjZC9P"
   },
   "source": [
    "## **Check Palindrome String**\n",
    "The word and its reverse spell out the same\n",
    "-> Racecar in reverse is Racecar"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "vUcrIFh3ZP53"
   },
   "source": [
    "**Recursion**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:12:08.522198Z",
     "start_time": "2021-09-16T02:12:08.522198Z"
    },
    "id": "s0jnQv0bZG4M"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "palindrome!\n"
     ]
    }
   ],
   "source": [
    "def is_palindrome(s):\n",
    "    if len(s) < 1:\n",
    "        return True\n",
    "    else:\n",
    "        if s[0] == s[-1]:\n",
    "            return is_palindrome(s[1:-1])\n",
    "        else:\n",
    "            return False\n",
    "        \n",
    "a=\"racecar\"\n",
    "\n",
    "if(is_palindrome(a)==True):\n",
    "    print(\"palindrome!\")\n",
    "else:\n",
    "    print(\"Not a palindrome!\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "-bJVHzKBZhJm"
   },
   "source": [
    "**Loop**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-09-16T02:12:08.524196Z",
     "start_time": "2021-09-16T02:12:08.524196Z"
    },
    "id": "P3a6w6-vZiVx"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Palindrome\n"
     ]
    }
   ],
   "source": [
    "def is_palindrome(string):\n",
    "    reversed_string=\"\"\n",
    "   # transversing through string from last\n",
    "    for i in range(len(string),0,-1):\n",
    "      # Addind last characters of string into a new string\n",
    "        reversed_string+=string[i-1]\n",
    "    if string==reversed_string:\n",
    "        print(\"Palindrome\")\n",
    "    else:\n",
    "        print(\"Not a palindrome\")\n",
    "            \n",
    "\n",
    "is_palindrome(\"racecar\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'darGpu'"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def mystery(string):\n",
    "    if len(string)==1:\n",
    "        return string\n",
    "    return string[-1]+mystery(string[:-1])\n",
    "\n",
    "\n",
    "mystery(\"upGrad\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "aa\n",
      "ab\n",
      "ba\n",
      "bb\n"
     ]
    }
   ],
   "source": [
    "def magic_print(values,string,k):\n",
    "    if k==0:\n",
    "        print(string)\n",
    "        return\n",
    "    for i in values:\n",
    "        magic_print(values,string+i,k-1)\n",
    "\n",
    "values= 'ab'\n",
    "count = 2\n",
    "magic_print(values,\"\",count)"
   ]
  }
 ],
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