724. Find Pivot Index - Explanation

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

Topics

Array Prefix Sum

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Prerequisites

Before attempting this problem, you should be comfortable with:

Prefix Sum - Computing cumulative sums to efficiently query range sums in O(1) time
Array Traversal - Iterating through arrays while maintaining running totals

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

Iterate through each index i from 0 to n-1.
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.
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

public class Solution {
    public int pivotIndex(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            int leftSum = 0, rightSum = 0;
            for (int l = 0; l < i; l++) {
                leftSum += nums[l];
            }
            for (int r = i + 1; r < n; r++) {
                rightSum += nums[r];
            }
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
}

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            int leftSum = 0, rightSum = 0;
            for (int l = 0; l < i; l++) {
                leftSum += nums[l];
            }
            for (int r = i + 1; r < n; r++) {
                rightSum += nums[r];
            }
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    pivotIndex(nums) {
        const n = nums.length;
        for (let i = 0; i < n; i++) {
            let leftSum = 0,
                rightSum = 0;
            for (let l = 0; l < i; l++) {
                leftSum += nums[l];
            }
            for (let r = i + 1; r < n; r++) {
                rightSum += nums[r];
            }
            if (leftSum === rightSum) {
                return i;
            }
        }
        return -1;
    }
}

public class Solution {
    public int PivotIndex(int[] nums) {
        int n = nums.Length;
        for (int i = 0; i < n; i++) {
            int leftSum = 0, rightSum = 0;
            for (int l = 0; l < i; l++) {
                leftSum += nums[l];
            }
            for (int r = i + 1; r < n; r++) {
                rightSum += nums[r];
            }
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
}

func pivotIndex(nums []int) int {
    n := len(nums)
    for i := 0; i < n; i++ {
        leftSum, rightSum := 0, 0
        for l := 0; l < i; l++ {
            leftSum += nums[l]
        }
        for r := i + 1; r < n; r++ {
            rightSum += nums[r]
        }
        if leftSum == rightSum {
            return i
        }
    }
    return -1
}

class Solution {
    fun pivotIndex(nums: IntArray): Int {
        val n = nums.size
        for (i in 0 until n) {
            var leftSum = 0
            var rightSum = 0
            for (l in 0 until i) {
                leftSum += nums[l]
            }
            for (r in i + 1 until n) {
                rightSum += nums[r]
            }
            if (leftSum == rightSum) {
                return i
            }
        }
        return -1
    }
}

class Solution {
    func pivotIndex(_ nums: [Int]) -> Int {
        let n = nums.count
        for i in 0..<n {
            var leftSum = 0
            var rightSum = 0
            for l in 0..<i {
                leftSum += nums[l]
            }
            for r in (i + 1)..<n {
                rightSum += nums[r]
            }
            if leftSum == rightSum {
                return i
            }
        }
        return -1
    }
}

impl Solution {
    pub fn pivot_index(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        for i in 0..n {
            let left_sum: i32 = nums[..i].iter().sum();
            let right_sum: i32 = nums[i + 1..].iter().sum();
            if left_sum == right_sum {
                return i as i32;
            }
        }
        -1
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$

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

Build a prefix sum array where prefixSum[i+1] = prefixSum[i] + nums[i].
For each index i:
- Left sum = prefixSum[i].
- Right sum = prefixSum[n] - prefixSum[i+1].
- If they are equal, return i.
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

public class Solution {
    public int pivotIndex(int[] nums) {
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        for (int i = 0; i < n; i++) {
            int leftSum = prefixSum[i];
            int rightSum = prefixSum[n] - prefixSum[i + 1];
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
}

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int n = nums.size();
        vector<int> prefixSum(n + 1, 0);
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        for (int i = 0; i < n; i++) {
            int leftSum = prefixSum[i];
            int rightSum = prefixSum[n] - prefixSum[i + 1];
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    pivotIndex(nums) {
        const n = nums.length;
        const prefixSum = new Array(n + 1).fill(0);
        for (let i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        for (let i = 0; i < n; i++) {
            const leftSum = prefixSum[i];
            const rightSum = prefixSum[n] - prefixSum[i + 1];
            if (leftSum === rightSum) {
                return i;
            }
        }
        return -1;
    }
}

public class Solution {
    public int PivotIndex(int[] nums) {
        int n = nums.Length;
        int[] prefixSum = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }
        for (int i = 0; i < n; i++) {
            int leftSum = prefixSum[i];
            int rightSum = prefixSum[n] - prefixSum[i + 1];
            if (leftSum == rightSum) {
                return i;
            }
        }
        return -1;
    }
}

class Solution {
    fun pivotIndex(nums: IntArray): Int {
        val n = nums.size
        val prefixSum = IntArray(n + 1)
        for (i in 0 until n) {
            prefixSum[i + 1] = prefixSum[i] + nums[i]
        }

        for (i in 0 until n) {
            val leftSum = prefixSum[i]
            val rightSum = prefixSum[n] - prefixSum[i + 1]
            if (leftSum == rightSum) {
                return i
            }
        }
        return -1
    }
}

class Solution {
    func pivotIndex(_ nums: [Int]) -> Int {
        let n = nums.count
        var prefixSum = [Int](repeating: 0, count: n + 1)
        for i in 0..<n {
            prefixSum[i + 1] = prefixSum[i] + nums[i]
        }

        for i in 0..<n {
            let leftSum = prefixSum[i]
            let rightSum = prefixSum[n] - prefixSum[i + 1]
            if leftSum == rightSum {
                return i
            }
        }
        return -1
    }
}

impl Solution {
    pub fn pivot_index(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut prefix_sum = vec![0; n + 1];
        for i in 0..n {
            prefix_sum[i + 1] = prefix_sum[i] + nums[i];
        }

        for i in 0..n {
            let left_sum = prefix_sum[i];
            let right_sum = prefix_sum[n] - prefix_sum[i + 1];
            if left_sum == right_sum {
                return i as i32;
            }
        }
        -1
    }
}

Time & Space Complexity

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

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

Compute the total sum of the array.
Initialize leftSum = 0.
For each index i:
- Compute rightSum = total - leftSum - nums[i].
- If leftSum == rightSum, return i.
- Add nums[i] to leftSum.
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

public class Solution {
    public int pivotIndex(int[] nums) {
        int total = 0;
        for (int num : nums) {
            total += num;
        }

        int leftSum = 0;
        for (int i = 0; i < nums.length; i++) {
            int rightSum = total - leftSum - nums[i];
            if (leftSum == rightSum) {
                return i;
            }
            leftSum += nums[i];
        }
        return -1;
    }
}

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int total = 0;
        for (int num : nums) {
            total += num;
        }

        int leftSum = 0;
        for (int i = 0; i < nums.size(); i++) {
            int rightSum = total - leftSum - nums[i];
            if (leftSum == rightSum) {
                return i;
            }
            leftSum += nums[i];
        }
        return -1;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    pivotIndex(nums) {
        let total = 0;
        for (let num of nums) {
            total += num;
        }

        let leftSum = 0;
        for (let i = 0; i < nums.length; i++) {
            let rightSum = total - leftSum - nums[i];
            if (leftSum === rightSum) {
                return i;
            }
            leftSum += nums[i];
        }
        return -1;
    }
}

public class Solution {
    public int PivotIndex(int[] nums) {
        int total = 0;
        foreach (int num in nums) {
            total += num;
        }
        int leftSum = 0;
        for (int i = 0; i < nums.Length; i++) {
            int rightSum = total - nums[i] - leftSum;
            if (leftSum == rightSum) {
                return i;
            }
            leftSum += nums[i];
        }
        return -1;
    }
}

func pivotIndex(nums []int) int {
    total := 0
    for _, num := range nums {
        total += num
    }

    leftSum := 0
    for i := 0; i < len(nums); i++ {
        rightSum := total - leftSum - nums[i]
        if leftSum == rightSum {
            return i
        }
        leftSum += nums[i]
    }
    return -1
}

class Solution {
    fun pivotIndex(nums: IntArray): Int {
        var total = 0
        for (num in nums) {
            total += num
        }

        var leftSum = 0
        for (i in nums.indices) {
            val rightSum = total - leftSum - nums[i]
            if (leftSum == rightSum) {
                return i
            }
            leftSum += nums[i]
        }
        return -1
    }
}

class Solution {
    func pivotIndex(_ nums: [Int]) -> Int {
        var total = 0
        for num in nums {
            total += num
        }

        var leftSum = 0
        for i in 0..<nums.count {
            let rightSum = total - leftSum - nums[i]
            if leftSum == rightSum {
                return i
            }
            leftSum += nums[i]
        }
        return -1
    }
}

impl Solution {
    pub fn pivot_index(nums: Vec<i32>) -> i32 {
        let total: i32 = nums.iter().sum();
        let mut left_sum = 0;
        for i in 0..nums.len() {
            let right_sum = total - left_sum - nums[i];
            if left_sum == right_sum {
                return i as i32;
            }
            left_sum += nums[i];
        }
        -1
    }
}

Time & Space Complexity

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

Common Pitfalls

Confusing Empty Sum as Zero

When the pivot is at index 0, the left sum is 0 (sum of no elements). Similarly, when the pivot is at the last index, the right sum is 0. Some solutions incorrectly skip checking the first or last index, assuming a pivot must have elements on both sides. The problem defines sums of empty ranges as 0, so boundary indices are valid pivot candidates.

Including the Pivot Element in the Sum

The pivot index itself should not be included in either the left sum or the right sum. A common mistake is computing leftSum as the sum up to and including index i, or computing rightSum starting from index i. The correct formula is rightSum = total - leftSum - nums[i], which explicitly excludes the pivot element from both sums.