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Problem solution · Python

Minimize the Maximum of Two Arrays

Minimize the Maximum of Two Arrays: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
33 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Minimize the Maximum of Two Arrays, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 33 lines of Python from the credited upstream file 2513.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimize the Maximum of Two Arrays · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minimizeSet(      self,      divisor1: int,      divisor2: int,      uniqueCnt1: int,      uniqueCnt2: int,  ) -> int:    divisorLcm = math.lcm(divisor1, divisor2)    l = 0    r = 2**31 - 1     def isPossible(m: int) -> bool:      """      Returns True if we can take uniqueCnt1 integers from [1..m] to arr1 and      take uniqueCnt2 integers from [1..m] to arr2.      """      cnt1 = m - m // divisor1      cnt2 = m - m // divisor2      totalCnt = m - m // divisorLcm      return (cnt1 >= uniqueCnt1 and              cnt2 >= uniqueCnt2 and              totalCnt >= uniqueCnt1 + uniqueCnt2)     while l < r:      m = (l + r) // 2      if isPossible(m):        r = m      else:        l = m + 1     return l

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