LeetCode in Kotlin

3424. Minimum Cost to Make Arrays Identical

Medium

You are given two integer arrays arr and brr of length n, and an integer k. You can perform the following operations on arr any number of times:

• Split arr into any number of contiguous non-empty subarrays and rearrange these subarrays in any order. This operation has a fixed cost of k.
• Choose any element in arr and add or subtract a positive integer x to it. The cost of this operation is x.

Return the minimum total cost to make arr equal to brr.

Example 1:

Input: arr = [-7,9,5], brr = [7,-2,-5], k = 2

Output: 13

Explanation:

• Split arr into two contiguous subarrays: [-7] and [9, 5] and rearrange them as [9, 5, -7], with a cost of 2.
• Subtract 2 from element arr[0]. The array becomes [7, 5, -7]. The cost of this operation is 2.
• Subtract 7 from element arr[1]. The array becomes [7, -2, -7]. The cost of this operation is 7.
• Add 2 to element arr[2]. The array becomes [7, -2, -5]. The cost of this operation is 2.

The total cost to make the arrays equal is 2 + 2 + 7 + 2 = 13.

Example 2:

Input: arr = [2,1], brr = [2,1], k = 0

Output: 0

Explanation:

Since the arrays are already equal, no operations are needed, and the total cost is 0.

Constraints:

• 1 <= arr.length == brr.length <= 105
• 0 <= k <= 2 * 1010
• -105 <= arr[i] <= 105
• -105 <= brr[i] <= 105

Solution

import kotlin.math.abs
import kotlin.math.min

class Solution {
    fun minCost(arr: IntArray, brr: IntArray, k: Long): Long {
        val n = arr.size
        var sum1: Long = 0
        var sum2: Long
        for (i in 0..<n) {
            sum1 = sum1 + abs((arr[i] - brr[i]))
        }
        if (k < sum1) {
            arr.sort()
            brr.sort()
            sum2 = k
            for (i in 0..<n) {
                sum2 = sum2 + abs((arr[i] - brr[i]))
            }
        } else {
            return sum1
        }
        return min(sum1, sum2)
    }
}

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