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Find Pivot Index

Problem​

Given an array of integers nums, calculate the pivot index of this array.

The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the right of the index.

If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.

Return the leftmost pivot index. If no such index exists, return -1.

Constraints:
  • 1 <= nums.length <= 10⁴
  • -1000 <= nums[i] <= 1000
Examples
Input: nums = [1,7,3,6,5,6]
Output: 3
Explanation:
The pivot index is 3.
Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11
Right sum = nums[4] + nums[5] = 5 + 6 = 11
Input: nums = [1,2,3]
Output: -1
Explanation: There is no index that satisfies the conditions in the problem statement.
Input: nums = [2,1,-1]
Output: 0
Explanation:
The pivot index is 0.
Left sum = 0 (no elements to the left of index 0)
Right sum = nums[1] + nums[2] = 1 + (-1) = 0

Solution​

This problem can be efficiently solved using the prefix sum technique. Instead of calculating left and right sums repeatedly for each index, we can compute them smartly in a single pass.

Key Insight:

  • For any index i, if leftSum is the sum of all elements to its left, and total is the sum of all elements in the array, then:
    • rightSum = total - leftSum - nums[i]
  • We want to find where leftSum === rightSum

Algorithm:

  1. Calculate the total sum of all elements in the array
  2. Initialize leftSum = 0
  3. Iterate through the array:
    • Calculate rightSum = total - leftSum - nums[i]
    • If leftSum === rightSum, return the current index i
    • Update leftSum by adding nums[i]
  4. If no pivot index is found, return -1

Example walkthrough:

For nums = [1, 7, 3, 6, 5, 6], total = 28:

i=0: leftSum=0, rightSum=28-0-1=27, 0≠27, leftSum becomes 1
i=1: leftSum=1, rightSum=28-1-7=20, 1≠20, leftSum becomes 8
i=2: leftSum=8, rightSum=28-8-3=17, 8≠17, leftSum becomes 11
i=3: leftSum=11, rightSum=28-11-6=11, 11===11 ✓ return 3

Time Complexity: O(n) - one pass to calculate total sum, one pass to find pivot Space Complexity: O(1) - only using constant extra space

function pivotIndex(nums: number[]): number {
  const len = nums.length
  const sum = nums.reduce((acc, val) => acc + val)
  let leftSum = 0

  for (let i = 0; i < len; i++) {
    if (sum - nums[i] - leftSum === leftSum) {
      return i
    }
    leftSum += nums[i]
  }

  return -1
}