2528. Maximize the Minimum Powered City

Description

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. In other words, if the range is denoted by r, then a power station at city i can provide power to all cities j such that |i - j| <= r and 0 <= i, j <= n - 1.

• Note that |x| denotes absolute value. For example, |7 - 5| = 2 and |3 - 10| = 7.

The power of a city is the total number of power stations it is being provided power from.

The government has sanctioned building k more power stations, each of which can be built in any city, and have the same range as the pre-existing ones.

Given the two integers r and k, return the maximum possible minimum power of a city, if the additional power stations are built optimally.

Note that you can build the k power stations in multiple cities.

 

Example 1:

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.

Example 2:

Input: stations = [4,4,4,4], r = 0, k = 3
Output: 4
Explanation: 
It can be proved that we cannot make the minimum power of a city greater than 4.

 

Constraints:

• n == stations.length
• 1 <= n <= 105
• 0 <= stations[i] <= 105
• 0 <= r <= n - 1
• 0 <= k <= 109

Solution

maximize-the-minimum-powered-city.py

class Solution:
    def maxPower(self, stations: List[int], r: int, k: int) -> int:
        N = len(stations)
        prefix = [0] * (N + 1)

        for i in range(N):
            left = max(0, i - r)
            right = min(N - 1, i + r)

            prefix[left] += stations[i]
            prefix[right + 1] -= stations[i]

        for i in range(1, N):
            prefix[i] += prefix[i - 1]

        def good(mid):
            stations = k
            build = [0] * (N + 1)

            for i in range(N):
                if i > 0:
                    build[i] += build[i - 1]

                curr = prefix[i] + build[i]

                if curr < mid:
                    need = mid - curr
                    stations -= need
                    if stations < 0:
                        return False

                    j = min(i + r + r, N - 1)
                    build[i] += need
                    build[j + 1] -= need

            return stations >= 0

        left, right = 0, 10 ** 18
        res = 0

        while left <= right:
            mid = left + (right - left) // 2

            if good(mid):
                res = mid
                left = mid + 1
            else:
                right = mid - 1

        return res