How do you solve Maximize the Minimum Powered City on LeetCode?

Maximize the Minimum Powered City (LeetCode #2528) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max(stations) + k)) time complexity and O(n) space complexity. Key insight: Understanding how power stations affect multiple cities is crucial.

What is the brute force solution for Maximize the Minimum Powered City?

The brute force approach for Maximize the Minimum Powered City is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible way to distribute the additional power stations and calculating the minimum power for each configuration. This is straightforward but inefficient, as it checks all combinations. The optimal Optimal Solution approach improves this to O(n log(max(stations) + k)).

What are the interview tips for Maximize the Minimum Powered City?

Explain your thought process clearly while solving. Consider edge cases, such as when k is zero. Practice binary search problems to get comfortable with the technique.

What are common mistakes when solving Maximize the Minimum Powered City?

Not considering the range of power stations correctly. Failing to optimize the distribution of additional power stations.

#2528

Maximize the Minimum Powered City

Hard

ArrayBinary SearchGreedyQueueSliding WindowPrefix SumBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max(stations) + k))
Space	O(1)	O(n)

💡

Intuition

Time O(n log(max(stations) + k))Space O(n)

The optimal approach uses binary search to find the maximum possible minimum power of any city after optimally distributing the additional power stations. This is efficient and leverages the properties of the problem.

⚙️

Algorithm

3 steps

1Step 1: Define a function to check if a given minimum power can be achieved by distributing the k power stations.
2Step 2: Use binary search to find the maximum minimum power by checking midpoints and adjusting the search range based on feasibility.
3Step 3: For each midpoint, calculate how many additional power stations are needed to ensure every city has at least that power.

solution.py29 lines

1# Full working Python code
2
3def canAchievePower(stations, r, k, target):
4    n = len(stations)
5    needed = 0
6    power = [0] * n
7    for i in range(n):
8        if power[i] < target:
9            add = target - power[i]
10            needed += add
11            if needed > k:
12                return False
13            for j in range(max(0, i - r), min(n, i + r + 1)):
14                power[j] += add
15    return True
16
17
18def maxMinPower(stations, r, k):
19    left, right = 0, max(stations) + k
20    while left < right:
21        mid = (left + right + 1) // 2
22        if canAchievePower(stations, r, k, mid):
23            left = mid
24        else:
25            right = mid - 1
26    return left
27
28# Example usage
29print(maxMinPower([1, 2, 4, 5, 0], 1, 2))

ℹ

Complexity note: The time complexity is O(n log(max(stations) + k)) because we perform a binary search over the possible power values and for each value, we check if it can be achieved in O(n) time.

1Understanding how power stations affect multiple cities is crucial.
2Binary search can efficiently narrow down the maximum minimum power.

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