Introduction

In this blog post, we will explore how to solve the LeetCode problem "1664" 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

You are given an integer array nums. You can choose exactly one index (0-indexed) and remove the element. Notice that the index of the elements may change after the removal. For example, if nums = [6,1,7,4,1]: Choosing to remove index 1 results in nums = [6,7,4,1]. Choosing to remove index 2 results in nums = [6,1,4,1]. Choosing to remove index 4 results in nums = [6,1,7,4]. An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values. Return the number of indices that you could choose such that after the removal, nums is fair. Example 1: Input: nums = [2,1,6,4] Output: 1 Explanation: Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair. Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair. Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair. Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair. There is 1 index that you can remove to make nums fair. Example 2: Input: nums = [1,1,1] Output: 3 Explanation: You can remove any index and the remaining array is fair. Example 3: Input: nums = [1,2,3] Output: 0 Explanation: You cannot make a fair array after removing any index. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Prefix Sums: Calculate prefix sums for even and odd indices separately. This allows us to efficiently compute the sums of even and odd indices after removing an element.
• Iterate and Check: Iterate through the array. For each index i, calculate the sums of even and odd indices of the array after removing the element at i using the precomputed prefix sums.

• Fairness Check: Check if the even sum equals the odd sum after removal. Increment a counter if it's fair.

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

Code

    def waysToMakeFair(nums):
    n = len(nums)
    even_prefix = [0] * (n + 1)
    odd_prefix = [0] * (n + 1)

    for i in range(n):
        if i % 2 == 0:
            even_prefix[i + 1] = even_prefix[i] + nums[i]
            odd_prefix[i + 1] = odd_prefix[i]
        else:
            odd_prefix[i + 1] = odd_prefix[i] + nums[i]
            even_prefix[i + 1] = even_prefix[i]

    count = 0
    for i in range(n):
        even_sum = even_prefix[i] + (odd_prefix[n] - odd_prefix[i + 1])
        odd_sum = odd_prefix[i] + (even_prefix[n] - even_prefix[i + 1])

        if even_sum == odd_sum:
            count += 1

    return count