README_EN.md

2574. Left and Right Sum Differences

中文文档

Description

Given a 0-indexed integer array nums, find a 0-indexed integer array answer where:

• answer.length == nums.length.
• answer[i] = |leftSum[i] - rightSum[i]|.

Where:

• leftSum[i] is the sum of elements to the left of the index i in the array nums. If there is no such element, leftSum[i] = 0.
• rightSum[i] is the sum of elements to the right of the index i in the array nums. If there is no such element, rightSum[i] = 0.

Return the array answer.

 

Example 1:

Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].

Example 2:

Input: nums = [1]
Output: [0]
Explanation: The array leftSum is [0] and the array rightSum is [0].
The array answer is [|0 - 0|] = [0].

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 105

Solutions

Python3

class Solution:
    def leftRigthDifference(self, nums: List[int]) -> List[int]:
        left, right = 0, sum(nums)
        ans = []
        for x in nums:
            right -= x
            ans.append(abs(left - right))
            left += x
        return ans

Java

class Solution {
    public int[] leftRigthDifference(int[] nums) {
        int left = 0, right = Arrays.stream(nums).sum();
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            right -= nums[i];
            ans[i] = Math.abs(left - right);
            left += nums[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> leftRigthDifference(vector<int>& nums) {
        int left = 0, right = accumulate(nums.begin(), nums.end(), 0);
        vector<int> ans;
        for (int& x : nums) {
            right -= x;
            ans.push_back(abs(left - right));
            left += x;
        }
        return ans;
    }
};

Go

func leftRigthDifference(nums []int) (ans []int) {
	var left, right int
	for _, x := range nums {
		right += x
	}
	for _, x := range nums {
		right -= x
		ans = append(ans, abs(left-right))
		left += x
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

function leftRigthDifference(nums: number[]): number[] {
    let left = 0,
        right = nums.reduce((a, b) => a + b);
    const ans: number[] = [];
    for (const x of nums) {
        right -= x;
        ans.push(Math.abs(left - right));
        left += x;
    }
    return ans;
}

function leftRigthDifference(nums: number[]): number[] {
    let left = 0;
    let right = nums.reduce((r, v) => r + v);
    return nums.map(v => {
        right -= v;
        const res = Math.abs(left - right);
        left += v;
        return res;
    });
}

Rust

impl Solution {
    pub fn left_rigth_difference(nums: Vec<i32>) -> Vec<i32> {
        let mut left = 0;
        let mut right = nums.iter().sum::<i32>();
        nums.iter()
            .map(|v| {
                right -= v;
                let res = (left - right).abs();
                left += v;
                res
            })
            .collect()
    }
}

impl Solution {
    pub fn left_right_difference(nums: Vec<i32>) -> Vec<i32> {
        let mut ans = vec![];

        for i in 0..nums.len() {

            let mut left = 0;
            for j in 0..i {
                left += nums[j];
            }

            let mut right = 0;
            for k in i + 1..nums.len() {
                right += nums[k];
            }

            ans.push((left - right).abs());
        }

        ans
    }
}

impl Solution {
    pub fn left_right_difference(nums: Vec<i32>) -> Vec<i32> {
        let mut left = 0;
        let mut right: i32 = nums.iter().sum();
        let mut ans = vec![];

        for &x in &nums {
            right -= x;
            ans.push((left - right).abs());
            left += x;
        }

        ans
    }
}

C

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* leftRigthDifference(int* nums, int numsSize, int* returnSize) {
    int left = 0;
    int right = 0;
    for (int i = 0; i < numsSize; i++) {
        right += nums[i];
    }
    int* ans = malloc(sizeof(int) * numsSize);
    for (int i = 0; i < numsSize; i++) {
        right -= nums[i];
        ans[i] = abs(left - right);
        left += nums[i];
    }
    *returnSize = numsSize;
    return ans;
}

...