Find Pivot Index LeetCode Solution

Problem – Find Pivot Index

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

Find Pivot Index LeetCode Solution in Python

class Solution(object):
    def pivotIndex(self, nums):
        # Time: O(n)
        # Space: O(1)
        left, right = 0, sum(nums)
        for index, num in enumerate(nums):
            right -= num
            if left == right:
                return index
            left += num
        return -1

Find Pivot Index LeetCode Solution in Java

class Solution {
    public int pivotIndex(int[] nums) {
        int total = 0, sum = 0
        for (int num : nums) total += num;
        for (int i = 0; i < nums.length; sum += nums[i++])
            if (sum * 2 == total - nums[i]) return i;
        return -1;  
    }
}

Find Pivot Index LeetCode Solution in C++

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int total = 0;
        for (int num : nums) total += num;
        int sum = 0;
        for (int i = 0; i < nums.size(); sum += nums[i++])
            if (sum * 2 == total - nums[i])
                return i;
        
        return -1;
    }
};

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