45 Days of Leetcode - Find Pivot Index - Day 3.1

Hey dev! In case you don't know, I've started #45daysofleetcode and in these 45 days i'll be writing about two problems every day.

The problem, how I solved it and a bit of detailed explanation.

I'm sharing this progress with you so that you too can learn a bit of different perspective and have new solutions so hit like to join this journey.

I've chosen the best list of problems from medium to hard. That made lots of developers crack the interview.

Are you excited?

Today's Problem

Name
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

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

My Approach

We will iterate through the array, and at each index, we check if the sum of the left side of the index is equal to the sum of the right side of the index.

Code

class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        total_sum, right_sum, left_sum = sum(nums), 0, 0
        for i in range(len(nums)):
            right_sum = total_sum - nums[i] - left_sum
            if left_sum == right_sum:
                return i
            left_sum += nums[i]
        return -1

Let me know if you've any questions or suggestions!.

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