README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2513.Minimize%20the%20Maximum%20of%20Two%20Arrays/README_EN.md	2302	Biweekly Contest 94 Q3	Math Binary Search Number Theory

2513. Minimize the Maximum of Two Arrays

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Description

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

Solutions

Solution 1

Python3

class Solution:
    def minimizeSet(
        self, divisor1: int, divisor2: int, uniqueCnt1: int, uniqueCnt2: int
    ) -> int:
        def f(x):
            cnt1 = x // divisor1 * (divisor1 - 1) + x % divisor1
            cnt2 = x // divisor2 * (divisor2 - 1) + x % divisor2
            cnt = x // divisor * (divisor - 1) + x % divisor
            return (
                cnt1 >= uniqueCnt1
                and cnt2 >= uniqueCnt2
                and cnt >= uniqueCnt1 + uniqueCnt2
            )

        divisor = lcm(divisor1, divisor2)
        return bisect_left(range(10**10), True, key=f)

Java

class Solution {
    public int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
        long divisor = lcm(divisor1, divisor2);
        long left = 1, right = 10000000000L;
        while (left < right) {
            long mid = (left + right) >> 1;
            long cnt1 = mid / divisor1 * (divisor1 - 1) + mid % divisor1;
            long cnt2 = mid / divisor2 * (divisor2 - 1) + mid % divisor2;
            long cnt = mid / divisor * (divisor - 1) + mid % divisor;
            if (cnt1 >= uniqueCnt1 && cnt2 >= uniqueCnt2 && cnt >= uniqueCnt1 + uniqueCnt2) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return (int) left;
    }

    private long lcm(int a, int b) {
        return (long) a * b / gcd(a, b);
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

C++

class Solution {
public:
    int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
        long left = 1, right = 1e10;
        long divisor = lcm((long) divisor1, (long) divisor2);
        while (left < right) {
            long mid = (left + right) >> 1;
            long cnt1 = mid / divisor1 * (divisor1 - 1) + mid % divisor1;
            long cnt2 = mid / divisor2 * (divisor2 - 1) + mid % divisor2;
            long cnt = mid / divisor * (divisor - 1) + mid % divisor;
            if (cnt1 >= uniqueCnt1 && cnt2 >= uniqueCnt2 && cnt >= uniqueCnt1 + uniqueCnt2) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func minimizeSet(divisor1 int, divisor2 int, uniqueCnt1 int, uniqueCnt2 int) int {
	divisor := lcm(divisor1, divisor2)
	left, right := 1, 10000000000
	for left < right {
		mid := (left + right) >> 1
		cnt1 := mid/divisor1*(divisor1-1) + mid%divisor1
		cnt2 := mid/divisor2*(divisor2-1) + mid%divisor2
		cnt := mid/divisor*(divisor-1) + mid%divisor
		if cnt1 >= uniqueCnt1 && cnt2 >= uniqueCnt2 && cnt >= uniqueCnt1+uniqueCnt2 {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}