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Problem solution · Python

Maximize the Minimum Powered City

Maximize the Minimum Powered City: 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
40 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 Maximize the Minimum Powered City, 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

  • 40 lines of Python from the credited upstream file 2528.py.
  • The implementation visibly relies on sequence storage.
  • 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 codeMaximize the Minimum Powered City · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def maxPower(self, stations: list[int], r: int, k: int) -> int:    n = len(stations)    left = min(stations)    right = sum(stations) + k + 1     def check(            stations: list[int],            additionalStations: int, minPower: int) -> bool:      """Returns True if each city can have at least `minPower`."""      # Initilaize `power` as the 0-th city's power - stations[r].      power = sum(stations[:r])       for i in range(n):        if i + r < n:          power += stations[i + r]  # `power` = sum(stations[i - r..i + r]).        if power < minPower:          requiredPower = minPower - power          # There're not enough stations to plant.          if requiredPower > additionalStations:            return False          # Greedily plant `requiredPower` power stations in the farthest place          # to cover as many cities as possible.          stations[min(n - 1, i + r)] += requiredPower          additionalStations -= requiredPower          power += requiredPower        if i - r >= 0:          power -= stations[i - r]       return True     while left < right:      mid = (left + right) // 2      if check(stations.copy(), k, mid):        left = mid + 1      else:        right = mid     return left - 1

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