Introduction

In this blog post, we will explore how to solve the LeetCode problem "724" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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/

Explanation

Here's a breakdown of the approach and the Python code:

• Calculate Total Sum: Compute the sum of all elements in the array. This will allow us to efficiently calculate the right sum as we iterate.
• Iterate and Compare: Iterate through the array, maintaining a running left_sum. For each index, the right_sum can be derived from the total_sum and left_sum. Check if left_sum == right_sum.

• Return Index or -1: If a pivot index is found, return it. If the loop completes without finding a pivot index, return -1.

• Runtime Complexity: O(n), where n is the length of the array.

• Storage Complexity: O(1).

Code

    def pivotIndex(nums: list[int]) -> int:
    """
    Finds the pivot index of an array.

    Args:
        nums: A list of integers.

    Returns:
        The leftmost pivot index, or -1 if no such index exists.
    """
    total_sum = sum(nums)
    left_sum = 0

    for i in range(len(nums)):
        right_sum = total_sum - left_sum - nums[i]
        if left_sum == right_sum:
            return i
        left_sum += nums[i]

    return -1