LeetCode in Kotlin

3724. Minimum Operations to Transform Array

Medium

You are given two integer arrays nums1 of length n and nums2 of length n + 1.

You want to transform nums1 into nums2 using the minimum number of operations.

You may perform the following operations any number of times, each time choosing an index i:

• Increase nums1[i] by 1.
• Decrease nums1[i] by 1.
• Append nums1[i] to the end of the array.

Return the minimum number of operations required to transform nums1 into nums2.

Example 1:

Input: nums1 = [2,8], nums2 = [1,7,3]

Output: 4

Explanation:

Step	i	Operation	nums1[i]	Updated nums1
1	0	Append	-	[2, 8, 2]
2	0	Decrement	Decreases to 1	[1, 8, 2]
3	1	Decrement	Decreases to 7	[1, 7, 2]
4	2	Increment	Increases to 3	[1, 7, 3]

Thus, after 4 operations nums1 is transformed into nums2.

Example 2:

Input: nums1 = [1,3,6], nums2 = [2,4,5,3]

Output: 4

Explanation:

Step	i	Operation	nums1[i]	Updated nums1
1	1	Append	-	[1, 3, 6, 3]
2	0	Increment	Increases to 2	[2, 3, 6, 3]
3	1	Increment	Increases to 4	[2, 4, 6, 3]
4	2	Decrement	Decreases to 5	[2, 4, 5, 3]

Thus, after 4 operations nums1 is transformed into nums2.

Example 3:

Input: nums1 = [2], nums2 = [3,4]

Output: 3

Explanation:

Step	i	Operation	nums1[i]	Updated nums1
1	0	Increment	Increases to 3	[3]
2	0	Append	-	[3, 3]
3	1	Increment	Increases to 4	[3, 4]

Thus, after 3 operations nums1 is transformed into nums2.

Constraints:

• 1 <= n == nums1.length <= 105
• nums2.length == n + 1
• 1 <= nums1[i], nums2[i] <= 105

Solution

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

class Solution {
    fun minOperations(nums1: IntArray, nums2: IntArray): Long {
        val n = nums1.size
        val last = nums2[n]
        var steps: Long = 1
        var minDiffFromLast = Long.MAX_VALUE
        for (i in 0..<n) {
            val min = min(nums1[i], nums2[i])
            val max = max(nums1[i], nums2[i])
            steps += abs(max - min).toLong()
            if (minDiffFromLast > 0) {
                if (min <= last && last <= max) {
                    minDiffFromLast = 0
                } else {
                    minDiffFromLast = min(
                        minDiffFromLast,
                        min(abs(min - last), abs(max - last)).toLong(),
                    )
                }
            }
        }
        return steps + minDiffFromLast
    }
}

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