#2513

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimize the Maximum of Two Arrays

Minimize the Maximum of Two Arrays requires finding the smallest possible maximum number across two arrays, using binary search over valid answer space.

math binary search number theory

Problem Statement

You are given two arrays, arr1 and arr2, which are initially empty. You need to add positive integers to them such that the following conditions are satisfied:

Given divisors divisor1, divisor2, and the required unique counts uniqueCnt1 and uniqueCnt2 for each array, return the minimum possible maximum integer that can be present in either array.

Examples

Example 1

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

Output: 4

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

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

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 Approach

Binary Search over Valid Answer Space

Use binary search over the valid range of possible maximum integers. This approach efficiently narrows down the smallest possible maximum by checking how values can be distributed into both arrays.

Mathematical Distribution Logic

For each candidate maximum value during binary search, calculate how many multiples of divisor1 and divisor2 can fit into each array while satisfying the required unique counts.

Optimization of Array Distribution

Adjust the numbers added to arr1 and arr2 to ensure both arrays meet their unique count conditions, optimizing the process to minimize the maximum integer in the final arrays.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the efficiency of the binary search and array distribution steps. The binary search reduces the search space, while the array calculation checks each candidate efficiently. Space complexity is dependent on the storage required for maintaining intermediate states during the search process.

What Interviewers Usually Probe

Candidate demonstrates strong understanding of binary search and optimization techniques.
Effective handling of divisor-related number theory, ensuring unique counts are respected.
Clear explanation of how to balance the distribution between the two arrays while minimizing the maximum.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the binary search boundaries could lead to unnecessary computation.
Incorrectly distributing the numbers between the arrays, causing an invalid configuration.
Failing to optimize the approach by prematurely narrowing the search space or miscalculating unique counts.

Follow-up variants

Change the divisors and unique counts to test different configurations.
Increase the range of possible numbers to test performance with larger constraints.
Modify the approach to handle more than two arrays, adding complexity to the distribution problem.

FAQ

What is the best way to approach the 'Minimize the Maximum of Two Arrays' problem?

The best approach is to use binary search over the possible maximum values, ensuring each candidate is validated through proper distribution of numbers into two arrays.

How does binary search work in this problem?

Binary search helps to efficiently narrow down the minimum maximum value by checking potential candidates and verifying if the conditions hold under that value.

What are common mistakes in solving the 'Minimize the Maximum of Two Arrays' problem?

Common mistakes include incorrectly managing the binary search bounds, failing to properly distribute numbers to satisfy unique counts, and miscalculating the maximum value.

Can this problem be solved without binary search?

It is possible, but binary search optimizes the process by narrowing down the possible maximum efficiently, making it the most practical solution.

What is the significance of divisor1 and divisor2 in this problem?

Divisors play a key role in how numbers are distributed into the arrays, ensuring that each array meets its unique count condition while maintaining the desired maximum value.

terminal

Solution

Solution 1

#### Python3

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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)

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