Leetcode #2528: Maximize the Minimum Powered City
In this guide, we solve Leetcode #2528 Maximize the Minimum Powered City in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array stations of length n, where stations[i] represents the number of power stations in the ith city. Each power station can provide power to every city in a fixed range.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Queue, Array, Binary Search, Prefix Sum, Sliding Window
Intuition
We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.
Expanding and shrinking the window lets us maintain validity without restarting the scan.
Approach
Grow the window with a right pointer, and shrink from the left only when the constraint is violated.
Track the best window as you go to keep the solution linear.
Steps:
- Expand the right end of the window.
- While invalid, move the left end to restore constraints.
- Update the best window found.
Example
Input: stations = [1,2,4,5,0], r = 1, k = 2
Output: 5
Explanation:
One of the optimal ways is to install both the power stations at city 1.
So stations will become [1,4,4,5,0].
- City 0 is provided by 1 + 4 = 5 power stations.
- City 1 is provided by 1 + 4 + 4 = 9 power stations.
- City 2 is provided by 4 + 4 + 5 = 13 power stations.
- City 3 is provided by 5 + 4 = 9 power stations.
- City 4 is provided by 5 + 0 = 5 power stations.
So the minimum power of a city is 5.
Since it is not possible to obtain a larger power, we return 5.
Python Solution
class Solution:
def maxPower(self, stations: List[int], r: int, k: int) -> int:
def check(x, k):
d = [0] * (n + 1)
t = 0
for i in range(n):
t += d[i]
dist = x - (s[i] + t)
if dist > 0:
if k < dist:
return False
k -= dist
j = min(i + r, n - 1)
left, right = max(0, j - r), min(j + r, n - 1)
d[left] += dist
d[right + 1] -= dist
t += dist
return True
n = len(stations)
d = [0] * (n + 1)
for i, v in enumerate(stations):
left, right = max(0, i - r), min(i + r, n - 1)
d[left] += v
d[right + 1] -= v
s = list(accumulate(d))
left, right = 0, 1 << 40
while left < right:
mid = (left + right + 1) >> 1
if check(mid, k):
left = mid
else:
right = mid - 1
return left
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.