Question

2574. Left and Right Sum Differences

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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:

• 0724. Find Pivot Index
• 1991. Find the Middle Index in Array

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

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;
  }
};

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()
  }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* leftRigthDifference(int* nums, int numsSize, int* returnSize) {
  int left = 0;
  int right = 0;
  for (int i = 0; i < numsSize; i++) {
    right += nums[i];
  }
  int* ans = malloc(sizeof(int) * numsSize);
  for (int i = 0; i < numsSize; i++) {
    right -= nums[i];
    ans[i] = abs(left - right);
    left += nums[i];
  }
  *returnSize = numsSize;
  return ans;
}

Solution 2

function leftRigthDifference(nums: number[]): number[] {
  let left = 0;
  let right = nums.reduce((r, v) => r + v);
  return nums.map(v => {
    right -= v;
    const res = Math.abs(left - right);
    left += v;
    return res;
  });
}

impl Solution {
  pub fn left_right_difference(nums: Vec<i32>) -> Vec<i32> {
    let mut ans = vec![];

    for i in 0..nums.len() {
      let mut left = 0;
      for j in 0..i {
        left += nums[j];
      }

      let mut right = 0;
      for k in i + 1..nums.len() {
        right += nums[k];
      }

      ans.push((left - right).abs());
    }

    ans
  }
}

Solution 3

impl Solution {
  pub fn left_right_difference(nums: Vec<i32>) -> Vec<i32> {
    let mut left = 0;
    let mut right: i32 = nums.iter().sum();
    let mut ans = vec![];

    for &x in &nums {
      right -= x;
      ans.push((left - right).abs());
      left += x;
    }

    ans
  }
}