Minimum Total Cost to Make Arrays Unequal

Problem Description

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.

Key Insights

• The goal is to ensure that no elements in nums1 match the corresponding elements in nums2 at the same index.
• Swapping elements incurs a cost based on the indices of the elements being swapped.
• If there are elements in nums1 that are identical to those in nums2 at the same indices, we need to find a way to replace them by swapping with elements that don’t cause new conflicts.
• The problem can potentially be unsolvable if there are too many identical pairs in both arrays.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can use a greedy approach along with a frequency count of the elements in both arrays. The algorithm involves the following steps:

1. Count the frequency of each number in both nums1 and nums2.
2. Identify the indices where nums1 has the same value as nums2.
3. Create a list of available numbers that can be used to replace the conflicting indices in nums1.
4. Swap elements in nums1 to resolve conflicts while keeping track of the total cost incurred from the swaps.
5. If at any point it is determined that the conflicts cannot be resolved, return -1.

The data structures used include arrays for storing the counts and lists for holding the indices of conflicts.

Code Solutions