Introduction

In this blog post, we will explore how to solve the LeetCode problem "2333" 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 positive 0-indexed integer arrays nums1 and nums2, both of length n. The sum of squared difference of arrays nums1 and nums2 is defined as the sum of (nums1[i] - nums2[i])2 for each 0 <= i < n. You are also given two positive integers k1 and k2. You can modify any of the elements of nums1 by +1 or -1 at most k1 times. Similarly, you can modify any of the elements of nums2 by +1 or -1 at most k2 times. Return the minimum sum of squared difference after modifying array nums1 at most k1 times and modifying array nums2 at most k2 times. Note: You are allowed to modify the array elements to become negative integers. Example 1: Input: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0 Output: 579 Explanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0. The sum of square difference will be: (1 - 2)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2 = 579. Example 2: Input: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1 Output: 43 Explanation: One way to obtain the minimum sum of square difference is: - Increase nums1[0] once. - Increase nums2[2] once. The minimum of the sum of square difference will be: (2 - 5)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2 = 43. Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43. Constraints: n == nums1.length == nums2.length 1 <= n <= 105 0 <= nums1[i], nums2[i] <= 105 0 <= k1, k2 <= 109

Explanation

Here's the breakdown of the solution:

• Focus on Differences: The problem is about minimizing the sum of squared differences. Instead of directly modifying nums1 and nums2, calculate the absolute differences between corresponding elements of the arrays. Then, optimally reduce these differences.
• Greedy Reduction: Apply a greedy strategy. Prioritize reducing the largest differences first, as squaring larger numbers has a much greater impact. Use a heap (priority queue) to efficiently find and reduce the maximum differences.

• Handle Remaining Operations: After processing all differences, it's possible to have remaining modification operations. If so, apply the remaining operations evenly to all differences.

• Runtime Complexity: O(n log n), where n is the length of the input arrays due to heap operations.

• Storage Complexity: O(n), primarily for storing the differences in the heap.

Code

    import heapq

def minSumSquareDiff(nums1, nums2, k1, k2):
    """
    Calculates the minimum sum of squared difference after modifying arrays.

    Args:
        nums1: The first array of integers.
        nums2: The second array of integers.
        k1: The maximum number of modifications allowed for nums1.
        k2: The maximum number of modifications allowed for nums2.

    Returns:
        The minimum sum of squared difference after modifications.
    """
    n = len(nums1)
    diffs = []
    for i in range(n):
        diff = abs(nums1[i] - nums2[i])
        heapq.heappush(diffs, -diff)  # Use negative values for max-heap

    total_ops = k1 + k2
    while total_ops > 0:
        max_diff = -heapq.heappop(diffs)
        if max_diff == 0:
            break

        max_diff -= 1
        heapq.heappush(diffs, -max_diff)
        total_ops -= 1

    sum_sq_diff = 0
    for diff in diffs:
        sum_sq_diff += (-diff) ** 2

    return sum_sq_diff