2574. Left and Right Sum Differences

Problem

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

Solution (Java)

class Solution {

    public int[] leftRightDifference(int[] nums) {
        if(nums != null && nums.length > 0) {
            int n = nums.length;
            int[] lsum = new int[n];
            int[] rsum = new int[n];
            for(int i = 1, j = n-2; i<n; i++,j--) {
                lsum[i] += lsum[i-1] + nums[i-1];
                rsum[j] += rsum[j+1] + nums[j+1];
            }
            for(int i =0;i<n;i++) {
                nums[i] = Math.abs(lsum[i] - rsum[i]);
            }
        }
        return nums;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).