LeetCode in Kotlin

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 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

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    private fun canIBeTheMinimum(power: LongArray, minimum: Long, k: Int, r: Int): Boolean {
        var k = k
        val n = power.size
        val extraPower = LongArray(n)
        for (i in 0 until n) {
            if (i > 0) {
                extraPower[i] += extraPower[i - 1]
            }
            val curPower = power[i] + extraPower[i]
            val req = minimum - curPower
            if (req <= 0) {
                continue
            }
            if (req > k) {
                return false
            }
            k -= req.toInt()
            extraPower[i] += req
            if (i + 2 * r + 1 < n) {
                extraPower[i + 2 * r + 1] -= req
            }
        }
        return true
    }

    private fun calculatePowerArray(stations: IntArray, r: Int): LongArray {
        val n = stations.size
        val preSum = LongArray(n)
        for (i in 0 until n) {
            var st = i - r
            val last = i + r + 1
            if (st < 0) {
                st = 0
            }
            preSum[st] += stations[i].toLong()
            if (last < n) {
                preSum[last] -= stations[i].toLong()
            }
        }
        for (i in 1 until n) {
            preSum[i] += preSum[i - 1]
        }
        return preSum
    }

    fun maxPower(stations: IntArray, r: Int, k: Int): Long {
        var min: Long = 0
        var sum = Math.pow(10.0, 10.0).toLong() + Math.pow(10.0, 9.0).toLong()
        val power = calculatePowerArray(stations, r)
        var ans: Long = -1
        while (min <= sum) {
            val mid = min + sum shr 1
            if (canIBeTheMinimum(power, mid, k, r)) {
                ans = mid
                min = mid + 1
            } else {
                sum = mid - 1
            }
        }
        return ans
    }
}

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