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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

Solutions

  • class Solution {
    private long[] s;
    private long[] d;
    private int n;

    public long maxPower(int[] stations, int r, int k) {
        n = stations.length;
        d = new long[n + 1];
        s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            int left = Math.max(0, i - r), right = Math.min(i + r, n - 1);
            d[left] += stations[i];
            d[right + 1] -= stations[i];
        }
        s[0] = d[0];
        for (int i = 1; i < n + 1; ++i) {
            s[i] = s[i - 1] + d[i];
        }
        long left = 0, right = 1l << 40;
        while (left < right) {
            long mid = (left + right + 1) >>> 1;
            if (check(mid, r, k)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private boolean check(long x, int r, int k) {
        Arrays.fill(d, 0);
        long t = 0;
        for (int i = 0; i < n; ++i) {
            t += d[i];
            long dist = x - (s[i] + t);
            if (dist > 0) {
                if (k < dist) {
                    return false;
                }
                k -= dist;
                int j = Math.min(i + r, n - 1);
                int left = Math.max(0, j - r), right = Math.min(j + r, n - 1);
                d[left] += dist;
                d[right + 1] -= dist;
                t += dist;
            }
        }
        return true;
    }
}

  • class Solution {
public:
    long long maxPower(vector<int>& stations, int r, int k) {
        int n = stations.size();
        long d[n + 1];
        memset(d, 0, sizeof d);
        for (int i = 0; i < n; ++i) {
            int left = max(0, i - r), right = min(i + r, n - 1);
            d[left] += stations[i];
            d[right + 1] -= stations[i];
        }
        long s[n + 1];
        s[0] = d[0];
        for (int i = 1; i < n + 1; ++i) {
            s[i] = s[i - 1] + d[i];
        }
        auto check = [&](long x, int k) {
            memset(d, 0, sizeof d);
            long t = 0;
            for (int i = 0; i < n; ++i) {
                t += d[i];
                long dist = x - (s[i] + t);
                if (dist > 0) {
                    if (k < dist) {
                        return false;
                    }
                    k -= dist;
                    int j = min(i + r, n - 1);
                    int left = max(0, j - r), right = min(j + r, n - 1);
                    d[left] += dist;
                    d[right + 1] -= dist;
                    t += dist;
                }
            }
            return true;
        };
        long left = 0, right = 1e12;
        while (left < right) {
            long mid = (left + right + 1) >> 1;
            if (check(mid, k)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
};

  • class Solution:
    def maxPower(self, stations: List[int], r: int, k: int) -> int:
        def check(x, k):
            d = [0] * (n + 1)
            t = 0
            for i in range(n):
                t += d[i]
                dist = x - (s[i] + t)
                if dist > 0:
                    if k < dist:
                        return False
                    k -= dist
                    j = min(i + r, n - 1)
                    left, right = max(0, j - r), min(j + r, n - 1)
                    d[left] += dist
                    d[right + 1] -= dist
                    t += dist
            return True

        n = len(stations)
        d = [0] * (n + 1)
        for i, v in enumerate(stations):
            left, right = max(0, i - r), min(i + r, n - 1)
            d[left] += v
            d[right + 1] -= v
        s = list(accumulate(d))
        left, right = 0, 1 << 40
        while left < right:
            mid = (left + right + 1) >> 1
            if check(mid, k):
                left = mid
            else:
                right = mid - 1
        return left


  • func maxPower(stations []int, r int, k int) int64 {
	n := len(stations)
	d := make([]int, n+1)
	s := make([]int, n+1)
	for i, v := range stations {
		left, right := max(0, i-r), min(i+r, n-1)
		d[left] += v
		d[right+1] -= v
	}
	s[0] = d[0]
	for i := 1; i < n+1; i++ {
		s[i] = s[i-1] + d[i]
	}
	check := func(x, k int) bool {
		d := make([]int, n+1)
		t := 0
		for i := range stations {
			t += d[i]
			dist := x - (s[i] + t)
			if dist > 0 {
				if k < dist {
					return false
				}
				k -= dist
				j := min(i+r, n-1)
				left, right := max(0, j-r), min(j+r, n-1)
				d[left] += dist
				d[right+1] -= dist
				t += dist
			}
		}
		return true
	}
	left, right := 0, 1<<40
	for left < right {
		mid := (left + right + 1) >> 1
		if check(mid, k) {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return int64(left)
}

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