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LeetCode 75 - 724. Find Pivot Index

Description

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:

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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:

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

Example 3:

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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

Solutions

Solution 1

This problem can be solved using a brute-force approach. By iterating through the array, we calculate the sum of elements on the left and right sides of each index and compare them to find the pivot index.

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class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int size = nums.size();
        int leftSum{};

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

Solution 2

This solution can be optimized based on the first approach. By using the prefix sum technique, we can avoid nested loops, reducing the time complexity from O(n^2) to O(n).

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