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# Group Anagrams (Medium) | ||
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## Table of Contents | ||
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- [Problem Statement](#problem-statement) | ||
- [Examples](#examples) | ||
- [Constraints](#constraints) | ||
- [Solutions](#solutions) | ||
- [Approach 1: Using Sorted Strings as Keys](#approach-1-using-sorted-strings-as-keys) | ||
- [Approach 2: Hash Map with Character Count as Key](#approach-2-hash-map-with-character-count-as-key) | ||
- [Complexity Analysis](#complexity-analysis) | ||
- [Code Explanation](#code-explanation) | ||
- [Related Resources](#related-resources) | ||
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## Problem Statement | ||
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Given an array of strings `strs`, group all anagrams together into sublists. You may return the output in any order. | ||
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An anagram is a string that contains the exact same characters as another string, but the order of the characters can be different. | ||
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- [View on Neetcode.io](https://neetcode.io/problems/anagram-groups) | ||
- [View on LeetCode](https://leetcode.com/problems/group-anagrams/description/) | ||
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## Examples | ||
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**Example 1:** | ||
``` | ||
Input: strs = ["eat","tea","tan","ate","nat","bat"] | ||
Output: [["bat"],["nat","tan"],["ate","eat","tea"]] | ||
``` | ||
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**Example 2:** | ||
``` | ||
Input: strs = [""] | ||
Output: [[""]] | ||
``` | ||
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**Example 3:** | ||
``` | ||
Input: strs = ["a"] | ||
Output: [["a"]] | ||
``` | ||
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## Constraints | ||
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- $1 \leq strs.length \leq 10^4$ | ||
- $0 \leq strs[i].length \leq 100$ | ||
- `strs[i]` consists of lowercase English letters. | ||
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## Solutions | ||
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### Approach 1: Using Sorted Strings as Keys | ||
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```python | ||
class Solution: | ||
def twoSum(self, nums: List[int], target: int) -> List[int]: | ||
if len(nums) == 2: | ||
return [0, 1] | ||
for i in range(len(nums)): | ||
for j in range(i + 1, len(nums)): | ||
if nums[i] + nums[j] == target: | ||
if nums.index(nums[i]) == nums.index(nums[j]): | ||
return [nums.index(nums[i]), nums.index(nums[j], i + 1, len(nums))] | ||
return [nums.index(nums[i]), nums.index(nums[j])] | ||
``` | ||
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### Approach 2: Hash Map with Character Count as Key | ||
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```python | ||
class Solution: | ||
def twoSum(self, nums: List[int], target: int) -> List[int]: | ||
seenMap = {} | ||
for i in range(len(nums)): | ||
remain = target - nums[i] | ||
if remain in seenMap: | ||
return [seenMap[remain], i] | ||
seenMap[nums[i]] = i | ||
return nums | ||
``` | ||
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## Complexity Analysis | ||
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### Approach 1: | ||
- Time Complexity: $O(N * M * \log(M))$, where N is the number of strings and M is the maximum length of a string. | ||
- Space Complexity: $O(N * M)$ | ||
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### Approach 2: | ||
- Time Complexity: $O(N * M)$, where N is the number of strings and M is the maximum length of a string. | ||
- Space Complexity: $O(N * M)$ | ||
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## Code Explanation | ||
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### Approach 1: Using Sorted Strings as Keys | ||
This approach involves sorting each string and using the sorted string as a key in a dictionary. | ||
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### Approach 2: Hash Map with Character Count as Key | ||
This solution uses a hash map to group anagrams: | ||
1. We iterate through each string in the input array. | ||
2. For each string, we create a key by counting the occurrences of each character. | ||
3. We use this count as a key in our hash map. The value is a list of all strings that have the same character count. | ||
4. Finally, we return the values of the hash map, which are the grouped anagrams. | ||
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This approach is efficient and passes on both Neetcode.io and LeetCode. | ||
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## Related Resources | ||
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- [Neetcode Solution](https://github.com/neetcode-gh/leetcode/blob/main/python/0049-group-anagrams.py) | ||
- [LeetCode Solution](https://leetcode.com/problems/group-anagrams/solutions/5729062/solution) | ||
- [Personal Submission](https://leetcode.com/submissions/detail/1377240624/) | ||
- [Video Explanation (Neetcode)](https://youtu.be/vzdNOK2oB2E?si=aFbSKVZz4WfzrNg1) | ||
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> [!NOTE] | ||
> This problem is part of a larger collection following the roadmap on [Neetcode Roadmap](https://neetcode.io/roadmap). For more details and related problems, please refer to the Neetcode.io website. |