Introduction

In this blog post, we will explore how to solve the LeetCode problem "2499" 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 0-indexed integer arrays nums1 and nums2, of equal length n. In one operation, you can swap the values of any two indices of nums1. The cost of this operation is the sum of the indices. Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n - 1 after performing all the operations. Return the minimum total cost such that nums1 and nums2 satisfy the above condition. In case it is not possible, return -1. Example 1: Input: nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5] Output: 10 Explanation: One of the ways we can perform the operations is: - Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5] - Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5]. - Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4]. We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10. Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10. Example 2: Input: nums1 = [2,2,2,1,3], nums2 = [1,2,2,3,3] Output: 10 Explanation: One of the ways we can perform the operations is: - Swap values at indices 2 and 3, incurring cost = 2 + 3 = 5. Now, nums1 = [2,2,1,2,3]. - Swap values at indices 1 and 4, incurring cost = 1 + 4 = 5. Now, nums1 = [2,3,1,2,2]. The total cost needed here is 10, which is the minimum possible. Example 3: Input: nums1 = [1,2,2], nums2 = [1,2,2] Output: -1 Explanation: It can be shown that it is not possible to satisfy the given conditions irrespective of the number of operations we perform. Hence, we return -1. Constraints: n == nums1.length == nums2.length 1 <= n <= 105 1 <= nums1[i], nums2[i] <= n

Explanation

Here's a breakdown of the solution approach, followed by the code:

• Identify Disagreements: Find indices where nums1[i] == nums2[i]. These are the indices that need to be swapped around.
• Cycle Decomposition: The core idea is to represent swaps as cycles. For example, if we need to move nums1[i] to nums1[j], and nums1[j] to nums1[k], and nums1[k] back to nums1[i], this forms a cycle. The cost of a cycle can be calculated directly.

• Minimum Cost Calculation: For each disagreement index, follow the swaps that need to be made to break the matches using cycle decomposition. Accumulate the sum of indices in the cycles. If it's impossible to resolve all disagreements, return -1.

• Runtime Complexity: O(N), where N is the length of the arrays.

• Storage Complexity: O(N)

Code

    def min_total_cost(nums1, nums2):
    n = len(nums1)
    mismatches = []
    for i in range(n):
        if nums1[i] == nums2[i]:
            mismatches.append(i)

    if not mismatches:
        return 0

    visited = [False] * n
    total_cost = 0

    for start in mismatches:
        if visited[start]:
            continue

        cycle = []
        curr = start
        while not visited[curr]:
            visited[curr] = True
            cycle.append(curr)

            next_val = nums1[curr]
            next_idx = -1
            for j in mismatches:
                if not visited[j] and nums2[j] == next_val:
                    next_idx = j
                    break

            if next_idx == -1:

                found_match = False
                for j in range(n):
                    if nums1[j] == nums2[j] and nums2[j] == next_val:
                        found_match = True
                        break
                if found_match:
                    return -1
                else:

                    for j in mismatches:
                        if nums2[j] == next_val:
                            next_idx = j
                            break
                    if next_idx == -1:


                        found_val = False
                        for val in nums1:
                            if val == next_val:
                                found_val = True
                                break
                        if not found_val:
                            return -1
                        else:
                           return -1

            curr = next_idx

        cycle_cost = 0
        for i in range(len(cycle)):
            cycle_cost += cycle[i]

        total_cost += cycle_cost

    num_mismatches = 0
    for i in range(n):
        if nums1[i] == nums2[i]:
            num_mismatches += 1

    if num_mismatches > 0:
        return -1

    return total_cost