最大化城市的最小电量

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

/*
题目思路：
1. 使用 Java Stream API 实现题解。
2. 通过将数组转换为流进行处理。
3. 计算每个城市在给定范围内的电量。
4. 使用二分搜索找到最小的最大电量。
*/
import java.util.Arrays;
import java.util.stream.IntStream;
public class Solution {
    public int maxPower(int[] stations, int r, int k) {
        int n = stations.length;
        long left = 0, right = (long) 1e18;
        while (left < right) {
            long mid = (left + right) / 2;
            if (canAchieve(stations, r, k, mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return (int) left;
    }
    private boolean canAchieve(int[] stations, int r, int k, long target) {
        int n = stations.length;
        long[] power = Arrays.stream(stations).mapToLong(i -> i).toArray();
        long additional = 0;
        for (int i = 0; i < n; i++) {
            if (i > 0) power[i] += power[i - 1];
            if (i >= 2 * r + 1) power[i] -= stations[i - 2 * r - 1];
            if (power[i] < target) {
                long add = target - power[i];
                additional += add;
                if (additional > k) return false;
                power[i] += add;
                if (i + r + 1 < n) stations[i + r + 1] += add;
            }
        }
        return true;
    }
}

题解采用了二分查找和差分数组的组合技术来解决问题。首先，根据电站的覆盖范围 r，初始化一个前缀和数组 pre，用于计算每个城市的初始供电情况。如果 r 覆盖整个数组，则简单地将 k 加到所有电站数之和上。否则，使用滑动窗口计算每个城市由覆盖范围内的电站提供的电力总和。接下来，使用二分搜索尝试找到可能的最小电量的最大值。对于二分搜索中的每个中间值 mid，使用差分数组来模拟在不同位置增加 k 个电站后的电力分布，以检查是否所有城市的电量都可以达到 mid。最终通过调整二分搜索的上下界，找到最大的可行电量。

O(n log(maxPower))

O(n)