Introduction

In this blog post, we will explore how to solve the LeetCode problem "2541" 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 two integer arrays nums1 and nums2 of equal length n and an integer k. You can perform the following operation on nums1: Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k. In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k. nums1 is said to be equal to nums2 if for all indices i such that 0 <= i < n, nums1[i] == nums2[i]. Return the minimum number of operations required to make nums1 equal to nums2. If it is impossible to make them equal, return -1. Example 1: Input: nums1 = [4,3,1,4], nums2 = [1,3,7,1], k = 3 Output: 2 Explanation: In 2 operations, we can transform nums1 to nums2. 1st operation: i = 2, j = 0. After applying the operation, nums1 = [1,3,4,4]. 2nd operation: i = 2, j = 3. After applying the operation, nums1 = [1,3,7,1]. One can prove that it is impossible to make arrays equal in fewer operations. Example 2: Input: nums1 = [3,8,5,2], nums2 = [2,4,1,6], k = 1 Output: -1 Explanation: It can be proved that it is impossible to make the two arrays equal. Constraints: n == nums1.length == nums2.length 2 <= n <= 105 0 <= nums1[i], nums2[j] <= 109 0 <= k <= 105

Explanation

Here's the breakdown of the solution:

• Check feasibility: The fundamental idea is that the sum of nums1 must be transformable to the sum of nums2 using only increments and decrements of k. This means the difference between their sums must be a multiple of k.
• Count increments and decrements: Iterate through the arrays and track the net difference nums2[i] - nums1[i] for each index. If this difference is not divisible by k, it's impossible to make the arrays equal. Maintain separate sums of positive differences (increments needed) and negative differences (decrements needed), where the amounts are divided by k.

• Return operations: The number of operations is simply the sum of the absolute values of differences divided by k.

• Time and Space Complexity: O(n) time complexity, O(1) space complexity

Code

    def minOperations(nums1, nums2, k):
    """
    Calculates the minimum number of operations to make nums1 equal to nums2.

    Args:
        nums1: The first integer array.
        nums2: The second integer array.
        k: The increment/decrement value.

    Returns:
        The minimum number of operations, or -1 if it's impossible.
    """
    n = len(nums1)
    sum1 = sum(nums1)
    sum2 = sum(nums2)

    if (sum2 - sum1) % k != 0:
        return -1

    increments = 0
    decrements = 0
    operations = 0

    for i in range(n):
        diff = nums2[i] - nums1[i]

        if diff % k != 0:
            return -1

        ops = diff // k

        if ops > 0:
            increments += ops
        elif ops < 0:
            decrements -= ops

    if increments != decrements:
        return -1

    for i in range(n):
        diff = nums2[i] - nums1[i]
        operations += abs(diff)

    return operations // k