Pivot Index of an Array

easy array prefix sum

Problem

Find the leftmost index p such that the sum of all elements strictly to its left equals the sum of all elements strictly to its right. If no such index exists, return -1. The element at p itself is excluded from both sides.

Inputnums = [2, 4, 1, 7, 3, 5]

Output3

Left of index 3: 2 + 4 + 1 = 7. Right of index 3: 3 + 5 = 8. They are not equal at index 3 — but at no index do they match in this example, so the actual return is -1. (Try [1, 7, 3, 6, 5, 6] for a true pivot at index 3.)

nums (comma-separated)

def pivot_index(nums):
    total = sum(nums)
    left = 0
    for i, x in enumerate(nums):
        right = total - left - x
        if left == right:
            return i
        left += x
    return -1

function pivotIndex(nums) {
  let total = 0;
  for (const x of nums) total += x;
  let left = 0;
  for (let i = 0; i < nums.length; i++) {
    const right = total - left - nums[i];
    if (left === right) return i;
    left += nums[i];
  }
  return -1;
}

class Solution {
    public int pivotIndex(int[] nums) {
        int total = 0;
        for (int x : nums) total += x;
        int left = 0;
        for (int i = 0; i < nums.length; i++) {
            int right = total - left - nums[i];
            if (left == right) return i;
            left += nums[i];
        }
        return -1;
    }
}

int pivotIndex(vector<int>& nums) {
    int total = 0;
    for (int x : nums) total += x;
    int left = 0;
    for (int i = 0; i < (int)nums.size(); i++) {
        int right = total - left - nums[i];
        if (left == right) return i;
        left += nums[i];
    }
    return -1;
}

Time: O(n) Space: O(1)