LeetCode in Kotlin

2513. Minimize the Maximum of Two Arrays

Medium

We have two arrays arr1 and arr2 which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions:

Given divisor1, divisor2, uniqueCnt1, and uniqueCnt2, return the minimum possible maximum integer that can be present in either array.

Example 1:

Input: divisor1 = 2, divisor2 = 7, uniqueCnt1 = 1, uniqueCnt2 = 3

Output: 4

Explanation:

We can distribute the first 4 natural numbers into arr1 and arr2.

arr1 = [1] and arr2 = [2,3,4].

We can see that both arrays satisfy all the conditions.

Since the maximum value is 4, we return it.

Example 2:

Input: divisor1 = 3, divisor2 = 5, uniqueCnt1 = 2, uniqueCnt2 = 1

Output: 3

Explanation:

Here arr1 = [1,2], and arr2 = [3] satisfy all conditions.

Since the maximum value is 3, we return it.

Example 3:

Input: divisor1 = 2, divisor2 = 4, uniqueCnt1 = 8, uniqueCnt2 = 2

Output: 15

Explanation:

Here, the final possible arrays can be arr1 = [1,3,5,7,9,11,13,15], and arr2 = [2,6].

It can be shown that it is not possible to obtain a lower maximum satisfying all conditions.

Constraints:

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    private fun gcd(a: Int, b: Int): Int {
        return if (b == 0) {
            a
        } else gcd(b, a % b)
    }

    fun minimizeSet(divisor1: Int, divisor2: Int, uniqueCnt1: Int, uniqueCnt2: Int): Int {
        var divisor2 = divisor2
        var lo: Long = 1
        var hi = 10e10.toInt().toLong()
        if (divisor2 == 0) {
            divisor2 = 1
        }
        val lcm = divisor1.toLong() * divisor2.toLong() / gcd(divisor1, divisor2)
        while (lo < hi) {
            val mi = lo + (hi - lo) / 2
            val cnt1 = (mi - mi / divisor1).toInt()
            val cnt2 = (mi - mi / divisor2).toInt()
            val cntAll = (mi - mi / lcm).toInt()
            if (cnt1 < uniqueCnt1 || cnt2 < uniqueCnt2 || cntAll < uniqueCnt1 + uniqueCnt2) {
                lo = mi + 1
            } else {
                hi = mi
            }
        }
        return lo.toInt()
    }
}