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Medium
Array
Dynamic Programming
Backtracking

494. Target Sum

中文文档

Description

You are given an integer array nums and an integer target.

You want to build an expression out of nums by adding one of the symbols '+' and '-' before each integer in nums and then concatenate all the integers.

  • For example, if nums = [2, 1], you can add a '+' before 2 and a '-' before 1 and concatenate them to build the expression "+2-1".

Return the number of different expressions that you can build, which evaluates to target.

 

Example 1:

Input: nums = [1,1,1,1,1], target = 3
Output: 5
Explanation: There are 5 ways to assign symbols to make the sum of nums be target 3.
-1 + 1 + 1 + 1 + 1 = 3
+1 - 1 + 1 + 1 + 1 = 3
+1 + 1 - 1 + 1 + 1 = 3
+1 + 1 + 1 - 1 + 1 = 3
+1 + 1 + 1 + 1 - 1 = 3

Example 2:

Input: nums = [1], target = 1
Output: 1

 

Constraints:

  • 1 <= nums.length <= 20
  • 0 <= nums[i] <= 1000
  • 0 <= sum(nums[i]) <= 1000
  • -1000 <= target <= 1000

Solutions

Solution 1: Dynamic Programming

Let's denote the sum of all elements in the array $\textit{nums}$ as $s$, and the sum of elements to which we assign a negative sign as $x$. Therefore, the sum of elements with a positive sign is $s - x$. We have:

$$ (s - x) - x = \textit{target} \Rightarrow x = \frac{s - \textit{target}}{2} $$

Since $x \geq 0$ and $x$ must be an integer, it follows that $s \geq \textit{target}$ and $s - \textit{target}$ must be even. If these two conditions are not met, we directly return $0$.

Next, we can transform the problem into: selecting several elements from the array $\textit{nums}$ such that the sum of these elements equals $\frac{s - \textit{target}}{2}$. We are asked how many ways there are to make such a selection.

We can use dynamic programming to solve this problem. Define $f[i][j]$ as the number of ways to select several elements from the first $i$ elements of the array $\textit{nums}$ such that the sum of these elements equals $j$.

For $\textit{nums}[i - 1]$, we have two choices: to select or not to select. If we do not select $\textit{nums}[i - 1]$, then $f[i][j] = f[i - 1][j]$; if we do select $\textit{nums}[i - 1]$, then $f[i][j] = f[i - 1][j - \textit{nums}[i - 1]]$. Therefore, the state transition equation is:

$$ f[i][j] = f[i - 1][j] + f[i - 1][j - \textit{nums}[i - 1]] $$

This selection is based on the premise that $j \geq \textit{nums}[i - 1]$.

The final answer is $f[m][n]$, where $m$ is the length of the array $\textit{nums}$, and $n = \frac{s - \textit{target}}{2}$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$.

Python3

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        s = sum(nums)
        if s < target or (s - target) % 2:
            return 0
        m, n = len(nums), (s - target) // 2
        f = [[0] * (n + 1) for _ in range(m + 1)]
        f[0][0] = 1
        for i, x in enumerate(nums, 1):
            for j in range(n + 1):
                f[i][j] = f[i - 1][j]
                if j >= x:
                    f[i][j] += f[i - 1][j - x]
        return f[m][n]

Java

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int s = Arrays.stream(nums).sum();
        if (s < target || (s - target) % 2 != 0) {
            return 0;
        }
        int m = nums.length;
        int n = (s - target) / 2;
        int[][] f = new int[m + 1][n + 1];
        f[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= nums[i - 1]) {
                    f[i][j] += f[i - 1][j - nums[i - 1]];
                }
            }
        }
        return f[m][n];
    }
}

C++

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s < target || (s - target) % 2) {
            return 0;
        }
        int m = nums.size();
        int n = (s - target) / 2;
        int f[m + 1][n + 1];
        memset(f, 0, sizeof(f));
        f[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= nums[i - 1]) {
                    f[i][j] += f[i - 1][j - nums[i - 1]];
                }
            }
        }
        return f[m][n];
    }
};

Go

func findTargetSumWays(nums []int, target int) int {
	s := 0
	for _, x := range nums {
		s += x
	}
	if s < target || (s-target)%2 != 0 {
		return 0
	}
	m, n := len(nums), (s-target)/2
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	f[0][0] = 1
	for i := 1; i <= m; i++ {
		for j := 0; j <= n; j++ {
			f[i][j] = f[i-1][j]
			if j >= nums[i-1] {
				f[i][j] += f[i-1][j-nums[i-1]]
			}
		}
	}
	return f[m][n]
}

TypeScript

function findTargetSumWays(nums: number[], target: number): number {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s < target || (s - target) % 2) {
        return 0;
    }
    const [m, n] = [nums.length, ((s - target) / 2) | 0];
    const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i <= m; i++) {
        for (let j = 0; j <= n; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= nums[i - 1]) {
                f[i][j] += f[i - 1][j - nums[i - 1]];
            }
        }
    }
    return f[m][n];
}

Rust

impl Solution {
    pub fn find_target_sum_ways(nums: Vec<i32>, target: i32) -> i32 {
        let s: i32 = nums.iter().sum();
        if s < target || (s - target) % 2 != 0 {
            return 0;
        }
        let m = nums.len();
        let n = ((s - target) / 2) as usize;
        let mut f = vec![vec![0; n + 1]; m + 1];
        f[0][0] = 1;
        for i in 1..=m {
            for j in 0..=n {
                f[i][j] = f[i - 1][j];
                if j as i32 >= nums[i - 1] {
                    f[i][j] += f[i - 1][j - nums[i - 1] as usize];
                }
            }
        }
        f[m][n]
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var findTargetSumWays = function (nums, target) {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s < target || (s - target) % 2) {
        return 0;
    }
    const [m, n] = [nums.length, ((s - target) / 2) | 0];
    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i <= m; i++) {
        for (let j = 0; j <= n; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= nums[i - 1]) {
                f[i][j] += f[i - 1][j - nums[i - 1]];
            }
        }
    }
    return f[m][n];
};

Solution 2: Dynamic Programming (Space Optimization)

We can observe that in the state transition equation of Solution 1, the value of $f[i][j]$ is only related to $f[i - 1][j]$ and $f[i - 1][j - \textit{nums}[i - 1]]$. Therefore, we can eliminate the first dimension of the space and use only a one-dimensional array.

The time complexity is $O(m \times n)$, and the space complexity is $O(n)$.

Python3

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        s = sum(nums)
        if s < target or (s - target) % 2:
            return 0
        n = (s - target) // 2
        f = [0] * (n + 1)
        f[0] = 1
        for x in nums:
            for j in range(n, x - 1, -1):
                f[j] += f[j - x]
        return f[n]

Java

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int s = Arrays.stream(nums).sum();
        if (s < target || (s - target) % 2 != 0) {
            return 0;
        }
        int n = (s - target) / 2;
        int[] f = new int[n + 1];
        f[0] = 1;
        for (int num : nums) {
            for (int j = n; j >= num; --j) {
                f[j] += f[j - num];
            }
        }
        return f[n];
    }
}

C++

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s < target || (s - target) % 2) {
            return 0;
        }
        int n = (s - target) / 2;
        int f[n + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int x : nums) {
            for (int j = n; j >= x; --j) {
                f[j] += f[j - x];
            }
        }
        return f[n];
    }
};

Go

func findTargetSumWays(nums []int, target int) int {
	s := 0
	for _, x := range nums {
		s += x
	}
	if s < target || (s-target)%2 != 0 {
		return 0
	}
	n := (s - target) / 2
	f := make([]int, n+1)
	f[0] = 1
	for _, x := range nums {
		for j := n; j >= x; j-- {
			f[j] += f[j-x]
		}
	}
	return f[n]
}

TypeScript

function findTargetSumWays(nums: number[], target: number): number {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s < target || (s - target) % 2) {
        return 0;
    }
    const n = ((s - target) / 2) | 0;
    const f = Array(n + 1).fill(0);
    f[0] = 1;
    for (const x of nums) {
        for (let j = n; j >= x; j--) {
            f[j] += f[j - x];
        }
    }
    return f[n];
}

Rust

impl Solution {
    pub fn find_target_sum_ways(nums: Vec<i32>, target: i32) -> i32 {
        let s: i32 = nums.iter().sum();
        if s < target || (s - target) % 2 != 0 {
            return 0;
        }
        let n = ((s - target) / 2) as usize;
        let mut f = vec![0; n + 1];
        f[0] = 1;
        for x in nums {
            for j in (x as usize..=n).rev() {
                f[j] += f[j - x as usize];
            }
        }
        f[n]
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var findTargetSumWays = function (nums, target) {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s < target || (s - target) % 2) {
        return 0;
    }
    const n = (s - target) / 2;
    const f = Array(n + 1).fill(0);
    f[0] = 1;
    for (const x of nums) {
        for (let j = n; j >= x; j--) {
            f[j] += f[j - x];
        }
    }
    return f[n];
};