LeetCode #724 — EASY

Find Pivot Index

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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:
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

Example 2:

Input: nums = [1,2,3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.

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 <= 10⁴
-1000 <= nums[i] <= 1000

Note: This question is the same as 1991: https://leetcode.com/problems/find-the-middle-index-in-array/

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,7,3,6,5,6]

Example 2

[1,2,3]

Example 3

[2,1,-1]

Core Insight

What unlocks the optimal approach

Create an array sumLeft where sumLeft[i] is the sum of all the numbers to the left of index i.
Create an array sumRight where sumRight[i] is the sum of all the numbers to the right of index i.
For each index i, check if sumLeft[i] equals sumRight[i]. If so, return i. If no such i is found, return -1.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #724: Find Pivot Index
class Solution {
    public int pivotIndex(int[] nums) {
        int left = 0, right = Arrays.stream(nums).sum();
        for (int i = 0; i < nums.length; ++i) {
            right -= nums[i];
            if (left == right) {
                return i;
            }
            left += nums[i];
        }
        return -1;
    }
}

// Accepted solution for LeetCode #724: Find Pivot Index
func pivotIndex(nums []int) int {
	var left, right int
	for _, x := range nums {
		right += x
	}
	for i, x := range nums {
		right -= x
		if left == right {
			return i
		}
		left += x
	}
	return -1
}

# Accepted solution for LeetCode #724: Find Pivot Index
class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        left, right = 0, sum(nums)
        for i, x in enumerate(nums):
            right -= x
            if left == right:
                return i
            left += x
        return -1

// Accepted solution for LeetCode #724: Find Pivot Index
impl Solution {
    pub fn pivot_index(nums: Vec<i32>) -> i32 {
        let (mut left, mut right): (i32, i32) = (0, nums.iter().sum());
        for i in 0..nums.len() {
            right -= nums[i];
            if left == right {
                return i as i32;
            }
            left += nums[i];
        }
        -1
    }
}

// Accepted solution for LeetCode #724: Find Pivot Index
function pivotIndex(nums: number[]): number {
    let left = 0,
        right = nums.reduce((a, b) => a + b);
    for (let i = 0; i < nums.length; ++i) {
        right -= nums[i];
        if (left == right) {
            return i;
        }
        left += nums[i];
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.