Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2574.html

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.

Similar problems:

  • 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()
        }
    }
    
    

All Problems

All Solutions