LeetCode in Kotlin

1818. Minimum Absolute Sum Difference

Medium

You are given two positive integer arrays nums1 and nums2, both of length n.

The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed).

You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference.

Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 109 + 7.

|x| is defined as:

Example 1:

Input: nums1 = [1,7,5], nums2 = [2,3,5]

Output: 3

Explanation: There are two possible optimal solutions:

Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.

Example 2:

Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]

Output: 0

Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an absolute sum difference of 0.

Example 3:

Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]

Output: 20

Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7]. This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20

Constraints:

Solution

import kotlin.math.abs

class Solution {
    fun minAbsoluteSumDiff(nums1: IntArray, nums2: IntArray): Int {
        var min = Int.MAX_VALUE
        var max = Int.MIN_VALUE
        for (i in nums1.indices) {
            min = min.coerceAtMost(nums1[i].coerceAtMost(nums2[i]))
            max = max.coerceAtLeast(nums1[i].coerceAtLeast(nums2[i]))
        }
        val less = IntArray(max - min + 1)
        val more = IntArray(max - min + 1)
        less[0] = -max - 1
        more[more.size - 1] = max + 1 shl 1
        for (num in nums1) {
            less[num - min] = num
            more[num - min] = num
        }
        for (i in 1 until less.size) {
            if (less[i] == 0) {
                less[i] = less[i - 1]
            }
        }
        for (i in more.size - 2 downTo 0) {
            if (more[i] == 0) {
                more[i] = more[i + 1]
            }
        }
        var total = 0
        var preSave = 0
        for (i in nums1.indices) {
            val current = abs(nums1[i] - nums2[i])
            total += current
            val save = (
                current -
                    abs(less[nums2[i] - min] - nums2[i]).coerceAtMost(abs(more[nums2[i] - min] - nums2[i]))
                )
            if (save > preSave) {
                total = total + preSave - save
                preSave = save
            }
            total %= 1000000007
        }
        return total
    }
}