How do you solve Minimize the Maximum of Two Arrays on LeetCode?

Minimize the Maximum of Two Arrays (LeetCode #2513) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(max)) time complexity and O(1) space complexity. Key insight: Understanding how to count valid integers based on divisibility is crucial.

What is the brute force solution for Minimize the Maximum of Two Arrays?

The brute force approach for Minimize the Maximum of Two Arrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible integers and checking which can be added to the two arrays while satisfying the conditions. This is straightforward but inefficient as it checks each integer one by one. The optimal Optimal Solution approach improves this to O(log(max)).

What algorithm or data structure is used to solve Minimize the Maximum of Two Arrays?

Minimize the Maximum of Two Arrays uses the following concepts: Math, Binary Search, Number Theory. The recommended patterns to study are: Binary Search, Counting.

What are the interview tips for Minimize the Maximum of Two Arrays?

Always clarify the constraints and edge cases before jumping into coding. Think about how you can optimize your solution from the start, especially for large input sizes. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Minimize the Maximum of Two Arrays?

Not considering the overlap of integers between the two arrays. Failing to correctly implement the counting function for valid integers.

#2513

Minimize the Maximum of Two Arrays

Medium

MathBinary SearchNumber TheoryBinary SearchCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(max))
Space	O(1)	O(1)

💡

Intuition

Time O(log(max))Space O(1)

The optimal approach uses binary search to minimize the maximum integer that can be present in either array. By leveraging the properties of divisibility and counting valid integers, we can efficiently find the smallest maximum.

⚙️

Algorithm

5 steps

1Step 1: Define a function to count how many valid integers can be added to both arrays up to a given maximum value.
2Step 2: Use binary search to find the minimum possible maximum integer that satisfies the conditions for both arrays.
3Step 3: Set the search range from 1 to a high enough value (like 2 * (uniqueCnt1 + uniqueCnt2)).
4Step 4: For each mid-point in the binary search, check if the counts of valid integers meet the unique count requirements.
5Step 5: Adjust the search range based on whether the current mid-point is valid.

solution.py18 lines

1# Full working Python code
2
3def count_valid(max_num, divisor1, divisor2):
4    count1 = max_num // divisor1
5    count2 = max_num // divisor2
6    count_common = max_num // (divisor1 * divisor2)
7    return (max_num - count1) + (max_num - count2) - count_common
8
9def minMax(divisor1, divisor2, uniqueCnt1, uniqueCnt2):
10    left, right = 1, 2 * (uniqueCnt1 + uniqueCnt2)
11    while left < right:
12        mid = (left + right) // 2
13        if count_valid(mid, divisor1, divisor2) >= uniqueCnt1 + uniqueCnt2:
14            right = mid
15        else:
16            left = mid + 1
17    return left
18

ℹ

Complexity note: The complexity is O(log(max)) where max is the maximum number we are searching up to, due to the binary search. The counting function runs in constant time.

1Understanding how to count valid integers based on divisibility is crucial.
2Binary search can significantly reduce the time complexity for problems involving finding minimum or maximum values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.