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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2501_2600.s2528_maximize_the_minimum_powered_city;

// #Hard #Array #Greedy #Binary_Search #Prefix_Sum #Sliding_Window #Queue
// #2023_04_19_Time_51_ms_(77.59%)_Space_60_MB_(5.17%)

/**
 * 2528 - Maximize the Minimum Powered City\.
 *
 * Hard
 *
 * You are given a **0-indexed** integer array `stations` of length `n`, where `stations[i]` represents the number of power stations in the i^th 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
**/
public class Solution {
    private boolean canIBeTheMinimum(long[] power, long minimum, int k, int r) {
        int n = power.length;
        long[] extraPower = new long[n];
        for (int i = 0; i < n; i++) {
            if (i > 0) {
                extraPower[i] += extraPower[i - 1];
            }
            long curPower = power[i] + extraPower[i];
            long req = minimum - curPower;
            if (req <= 0) {
                continue;
            }
            if (req > k) {
                return false;
            }
            k -= req;
            extraPower[i] += (req);
            if (i + 2 * r + 1 < n) {
                extraPower[i + 2 * r + 1] -= (req);
            }
        }
        return true;
    }

    private long[] calculatePowerArray(int[] stations, int r) {
        int n = stations.length;
        long[] preSum = new long[n];
        for (int i = 0; i < n; i++) {
            int st = i - r;
            int last = i + r + 1;
            if (st < 0) {
                st = 0;
            }
            preSum[st] += stations[i];
            if (last < n) {
                preSum[last] -= stations[i];
            }
        }
        for (int i = 1; i < n; i++) {
            preSum[i] += preSum[i - 1];
        }
        return preSum;
    }

    public long maxPower(int[] stations, int r, int k) {
        long min = 0;
        long sum = (long) Math.pow(10, 10) + (long) Math.pow(10, 9);
        long[] power = calculatePowerArray(stations, r);
        long ans = -1;
        while (min <= sum) {
            long mid = (min + sum) >> 1;
            if (canIBeTheMinimum(power, mid, k, r)) {
                ans = mid;
                min = mid + 1;
            } else {
                sum = mid - 1;
            }
        }
        return ans;
    }
}