2513. Minimize the Maximum of Two Arrays

Problem Link

Description

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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:

• arr1 contains uniqueCnt1 distinct positive integers, each of which is not divisible by divisor1.
• arr2 contains uniqueCnt2 distinct positive integers, each of which is not divisible by divisor2.
• No integer is present in both arr1 and arr2.

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:

• 2 <= divisor1, divisor2 <= 105
• 1 <= uniqueCnt1, uniqueCnt2 < 109
• 2 <= uniqueCnt1 + uniqueCnt2 <= 109

Solution

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Python3

class Solution:
    def minimizeSet(self, d1: int, d2: int, a: int, b: int) -> int:
        left, right = 1, 10 ** 18
        l = lcm(d1, d2)
        
        def good(k):
            A = k - k // d1
            B = k - k // d2
            C = k - k // l
            
            return A >= a and B >= b and C >= a + b
            
        
        while left < right:
            mid = left + (right - left) // 2
            
            if good(mid):
                right = mid
            else:
                left = mid + 1
            
        return left