Question

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

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)

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);
  }
}

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;
  }
};

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)
}