# 2574. Left and Right Sum Differences

## 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

Solution 1: Prefix Sum

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

We iterate over the array nums. For the current number $x$ we are iterating over, we update $right = right - x$. At this point, $left$ and $right$ represent the sum of the elements to the left and right of index $i$ in the array nums, respectively. We add the absolute difference between $left$ and $right$ to the answer array ans, and then update $left = left + x$.

After the iteration is complete, we return the answer array ans.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of the array nums.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}


• 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()
}
}