2574

2574. Left and Right Sum Differences (Easy)

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

Related Topics:
Array, Prefix Sum

Similar Questions:

• Find Pivot Index (Easy)
• Find the Middle Index in Array (Easy)
• Find the Distinct Difference Array (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/left-and-right-sum-differences
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> leftRightDifference(vector<int>& A) {
        int N = A.size();
        vector<int> leftSum(N), rightSum(N);
        for (int i = 1; i < N; ++i) {
            leftSum[i] = leftSum[i - 1] + A[i - 1];
            rightSum[N - i - 1] = rightSum[N - i] + A[N - i];
        }
        for (int i = 0; i < N; ++i) A[i] = abs(leftSum[i] - rightSum[i]);
        return A;
    }
};