LeetCode in Kotlin

2499. Minimum Total Cost to Make Arrays Unequal

Hard

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

Solution

class Solution {
    fun minimumTotalCost(nums1: IntArray, nums2: IntArray): Long {
        val n = nums1.size
        val bucket = IntArray(n + 1)
        var max = 0
        var maxKey = -1
        var totalBucket = 0
        var cost: Long = 0
        for (i in 0 until n) {
            if (nums1[i] == nums2[i]) {
                if (++bucket[nums1[i]] > max) {
                    max = bucket[nums1[i]]
                    maxKey = nums1[i]
                }
                totalBucket++
                cost += i.toLong()
            }
        }
        val requiredBucket = 2 * max
        if (requiredBucket > n) {
            return -1
        }
        var lackBucket = requiredBucket - totalBucket
        var i = 0
        while (i < n && lackBucket > 0) {
            if (nums1[i] == maxKey || nums2[i] == maxKey || nums1[i] == nums2[i]) {
                i++
                continue
            }
            lackBucket--
            cost += i.toLong()
            i++
        }
        return if (lackBucket > 0) {
            -1
        } else cost
    }
}

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