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2574. Left and Right Sum Differences

Description

You are given a 0-indexed integer array nums of size n.

Define two arrays leftSum and rightSum 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 an integer array answer of size n where answer[i] = |leftSum[i] - rightSum[i]|.

Β 

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].

Β 

Constraints:

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

Solutions

Solution 1: Prefix Sum

We define a variable \(l\) to represent the sum of elements to the left of index \(i\) in the array \(\textit{nums}\), and a variable \(r\) to represent the sum of elements to the right of index \(i\) in the array \(\textit{nums}\). Initially, \(l = 0\), \(r = \sum_{i = 0}^{n - 1} \textit{nums}[i]\).

We traverse the array \(\textit{nums}\). For the current number \(x\), we update \(r = r - x\). At this point, \(l\) and \(r\) represent the sum of elements to the left and right of index \(i\) in the array \(\textit{nums}\), respectively. We add the absolute difference of \(l\) and \(r\) to the answer array \(\textit{ans}\), then update \(l = l + x\).

After the traversal, we return the answer array \(\textit{ans}\).

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\), not counting the space for the return value.

Similar problems:

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

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class Solution {
    public int[] leftRightDifference(int[] nums) {
        int l = 0, r = 0;
        for (int x : nums) {
            r += x;
        }
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            r -= nums[i];
            ans[i] = Math.abs(l - r);
            l += nums[i];
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> leftRightDifference(vector<int>& nums) {
        int l = 0, r = 0;
        for (int x : nums) {
            r += x;
        }
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            r -= nums[i];
            ans[i] = abs(l - r);
            l += nums[i];
        }
        return ans;
    }
};

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func leftRightDifference(nums []int) []int {
    l, r := 0, 0
    for _, x := range nums {
        r += x
    }
    n := len(nums)
    ans := make([]int, n)
    for i, x := range nums {
        r -= x
        ans[i] = abs(l - r)
        l += x
    }
    return ans
}

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

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

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impl Solution {
    pub fn left_right_difference(nums: Vec<i32>) -> Vec<i32> {
        let mut l = 0;
        let mut r: i32 = nums.iter().sum();
        let mut ans = Vec::with_capacity(nums.len());
        for x in nums {
            r -= x;
            ans.push((l - r).abs());
            l += x;
        }
        ans
    }
}

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* leftRightDifference(int* nums, int numsSize, int* returnSize) {
    *returnSize = numsSize;
    int* ans = (int*) malloc(sizeof(int) * numsSize);

    int l = 0, r = 0;
    for (int i = 0; i < numsSize; ++i) {
        r += nums[i];
    }

    for (int i = 0; i < numsSize; ++i) {
        r -= nums[i];
        ans[i] = abs(l - r);
        l += nums[i];
    }

    return ans;
}

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