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.length1 <= n <= 1051 <= nums1[i], nums2[i] <= nThis approach uses a greedy strategy to minimize the total cost. We can consider swapping elements in nums1 to ensure that for every i, nums1[i] is not equal to nums2[i]. Here, we try to choose swaps that incur the minimum cost at each step, prioritizing swaps with the minimal sum of indices.
The Python code above iteratively tries to swap every pair of indices i and j in nums1 and checks whether the condition nums1[i] != nums2[i] is satisfied for all i. If it finds a swap that satisfies this condition, it returns the accumulated cost. If no such swap can be found at the end, it returns -1.
JavaScript
Time Complexity: O(n^2), as we can have nested loops generating pairs of indices to swap.
Space Complexity: O(1), as we use no additional space apart from tracking index pairs for minimum swaps.
In this approach, a two-pointers technique is employed to traverse both arrays, parallelly finding matches and performing swaps if necessary. This helps in maintaining synchronization and efficiently managing swaps to minimal encounters.
This C++ solution initializes two pointers - one from the start and one from the end - to traverse nums1. It performs swaps when nums1[i] aligns with nums2[i], borrowing elements from the end of the array to reduce incurred costs, thus optimizing swaps.
Java
Time Complexity: O(n), owing to the singular pass over nums1.
Space Complexity: O(1), given the constant space needed beyond input storage.
| Approach | Complexity |
|---|---|
| Greedy Approach | Time Complexity: O(n^2), as we can have nested loops generating pairs of indices to swap. |
| Two Pointers Approach | Time Complexity: O(n), owing to the singular pass over nums1. |
I solved too many Leetcode problems • NeetCodeIO • 101,495 views views
Watch 9 more video solutions →Practice Minimum Total Cost to Make Arrays Unequal with our built-in code editor and test cases.
Practice on FleetCode