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

Solution: Binary Search + Sliding Window + Difference Array

• Time complexity: O(nlog(sum(stations)+k))
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} stations
 * @param {number} r
 * @param {number} k
 * @return {number}
 */
const maxPower = function (stations, r, k) {
  const n = stations.length;
  const prefixSum = Array.from({ length: n + 1 }, () => 0);
  let left = Math.min(...stations);
  let right = stations.reduce((a, b) => a + b, 0) + k;

  for (let index = 1; index <= n; index++) {
    prefixSum[index] = prefixSum[index - 1] + stations[index - 1];
  }

  const isValidPower = target => {
    const diffs = new Array(n + 2).fill(0);
    let remainStations = k;

    for (let index = 1; index <= n; index++) {
      diffs[index] += diffs[index - 1];

      const left = Math.max(0, index - r - 1);
      const right = Math.min(n, index + r);
      const powers = prefixSum[right] - prefixSum[left] + diffs[index];

      if (powers < target) {
        const needStations = target - powers;

        if (needStations > remainStations) return false;

        const edge = Math.min(n + 1, index + r * 2 + 1);

        remainStations -= needStations;
        diffs[index] += needStations;
        diffs[edge] -= needStations;
      }
    }

    return true;
  };

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);

    isValidPower(mid) ? (left = mid + 1) : (right = mid - 1);
  }

  return right;
};