LeetCode in Kotlin

2333. Minimum Sum of Squared Difference

Medium

You are given two positive 0-indexed integer arrays nums1 and nums2, both of length n.

The sum of squared difference of arrays nums1 and nums2 is defined as the sum of (nums1[i] - nums2[i])2 for each 0 <= i < n.

You are also given two positive integers k1 and k2. You can modify any of the elements of nums1 by +1 or -1 at most k1 times. Similarly, you can modify any of the elements of nums2 by +1 or -1 at most k2 times.

Return the minimum sum of squared difference after modifying array nums1 at most k1 times and modifying array nums2 at most k2 times.

Note: You are allowed to modify the array elements to become negative integers.

Example 1:

Input: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0

Output: 579

Explanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0.

The sum of square difference will be: (1 - 2)² + (2 - 10)² + (3 - 20)² + (4 - 19)² = 579.

Example 2:

Input: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1

Output: 43

Explanation: One way to obtain the minimum sum of square difference is:

• Increase nums1[0] once.

• Increase nums2[2] once.

The minimum of the sum of square difference will be: (2 - 5)² + (4 - 8)² + (10 - 7)² + (12 - 9)² = 43.

Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.

Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 0 <= nums1[i], nums2[i] <= 105
• 0 <= k1, k2 <= 109

Solution

import kotlin.math.abs
import kotlin.math.pow

class Solution {
    fun minSumSquareDiff(nums1: IntArray, nums2: IntArray, k1: Int, k2: Int): Long {
        var minSumSquare: Long = 0
        val diffs = IntArray(100001)
        var totalDiff: Long = 0
        var kSum = k1.toLong() + k2
        var currentDiff: Int
        var maxDiff = 0
        for (i in nums1.indices) {
            // get current diff.
            currentDiff = abs(nums1[i] - nums2[i])
            // if current diff > 0, count/store it. If not,then ignore it.
            if (currentDiff > 0) {
                totalDiff += currentDiff.toLong()
                diffs[currentDiff]++
                maxDiff = maxDiff.coerceAtLeast(currentDiff)
            }
        }
        // if kSum (k1 + k2) < totalDifferences, it means we can make all numbers/differences 0s
        if (totalDiff <= kSum) {
            return 0
        }
        // starting from the back, from the highest difference, lower that group one by one to the
        // previous group.
        // we need to make all n diffs to n-1, then n-2, as long as kSum allows it.
        run {
            var i = maxDiff
            while (i > 0 && kSum > 0) {
                if (diffs[i] > 0) {
                    // if current group has more differences than the totalK, we can only move k of them
                    // to the lower level.
                    if (diffs[i] >= kSum) {
                        diffs[i] -= kSum.toInt()
                        diffs[i - 1] += kSum.toInt()
                        kSum = 0
                    } else {
                        // else, we can make this whole group one level lower.
                        diffs[i - 1] += diffs[i]
                        kSum -= diffs[i].toLong()
                        diffs[i] = 0
                    }
                }
                i--
            }
        }
        for (i in 0..maxDiff) {
            if (diffs[i] > 0) {
                minSumSquare += i.toDouble().pow(2.0).toLong() * diffs[i]
            }
        }
        return minSumSquare
    }
}

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