2528. Maximize the Minimum Powered City

Problem

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&nbsp;<= n - 1

• 0 <= k&nbsp;<= 109

Solution (Java)

class Solution {
    public long maxPower(int[] stations, int r, int k) {
        int n = stations.length;
        long left = 0, right = k;
        for (int x: stations)
            right += x;
        // v is the stations after adding
        long []v = new long[n];
        while (left <= right) {
            long x = (left + right) / 2;
            for (int i = 0; i < n; ++i) 
                v[i] = stations[i];
            // s means the power of city i
            // at first, it record the sum of v[0,r)
            long s = 0, use = 0;
            for (int i = 0; i < r; ++i) 
                s += v[i];
            for (int i = 0; i < n; ++i) {
                // add to t if needed
                int t = Math.min(n - 1, i + r);
                // update s
                // find a city should be added
                if (i + r < n) s += v[i + r];
                // find a city should be removed
                if (i - r > 0) s -= v[i - r - 1];
                // mising power stations
                long diff = Math.max(0, x - s);
                v[t] += diff;
                s += diff;
                use += diff;
            }
            if (use <= k) left = x + 1;
            else right = x - 1;
        }
        return right;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).