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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() } }

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