724. Find Pivot Index - Explanation

Problem Link

Description

You are 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 index's right.

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.

Example 1:

Input: nums = [1,7,3,6,5,6]

Output: 3

Explanation: Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11
Right sum = nums[4] + nums[5] = 5 + 6 = 11

Example 2:

Input: nums = [3,2,1]

Output: -1

Example 3:

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

Constraints:

  • 1 <= nums.length <= 10,000
  • -1,000 <= nums[i] <= 1,000


Company Tags

Please upgrade to NeetCode Pro to view company tags.



1. Brute Force

Intuition

The pivot index is where the sum of elements to the left equals the sum of elements to the right. The most straightforward approach is to check each index by computing both sums from scratch. For every potential pivot, sum all elements before it and all elements after it, then compare.

Algorithm

  1. Iterate through each index i from 0 to n-1.
  2. For each index:
    • Compute the left sum by adding all elements before index i.
    • Compute the right sum by adding all elements after index i.
    • If the two sums are equal, return i.
  3. If no pivot is found, return -1.
class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        n = len(nums)
        for i in range(n):
            leftSum = rightSum = 0
            for l in range(i):
                leftSum += nums[l]
            for r in range(i + 1, n):
                rightSum += nums[r]
            if leftSum == rightSum:
                return i
        return -1

Time & Space Complexity

  • Time complexity: O(n2)O(n ^ 2)
  • Space complexity: O(1)O(1)

2. Prefix Sum

Intuition

We can avoid recomputing sums repeatedly by precomputing a prefix sum array. The prefix sum at index i represents the sum of all elements from index 0 to i-1. With this, the left sum at any index is simply prefixSum[i], and the right sum is prefixSum[n] - prefixSum[i+1]. This reduces each lookup to constant time.

Algorithm

  1. Build a prefix sum array where prefixSum[i+1] = prefixSum[i] + nums[i].
  2. For each index i:
    • Left sum = prefixSum[i].
    • Right sum = prefixSum[n] - prefixSum[i+1].
    • If they are equal, return i.
  3. If no pivot is found, return -1.
class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        n = len(nums)
        prefixSum = [0] * (n + 1)
        for i in range(n):
            prefixSum[i + 1] = prefixSum[i] + nums[i]

        for i in range(n):
            leftSum = prefixSum[i]
            rightSum = prefixSum[n] - prefixSum[i + 1]
            if leftSum == rightSum:
                return i
        return -1

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(n)O(n)

3. Prefix Sum (Optimal)

Intuition

We can eliminate the need for a separate prefix sum array by maintaining a running left sum and computing the right sum on the fly. First, calculate the total sum of the array. As we iterate, the right sum at any index equals total - leftSum - nums[i]. We update leftSum after each comparison, keeping space usage constant.

Algorithm

  1. Compute the total sum of the array.
  2. Initialize leftSum = 0.
  3. For each index i:
    • Compute rightSum = total - leftSum - nums[i].
    • If leftSum == rightSum, return i.
    • Add nums[i] to leftSum.
  4. If no pivot is found, return -1.
class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        total = sum(nums)
        leftSum = 0
        for i in range(len(nums)):
            rightSum = total - nums[i] - leftSum
            if leftSum == rightSum:
                return i
            leftSum += nums[i]
        return -1

Time & Space Complexity

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