LeetCode in Kotlin

2071. Maximum Number of Tasks You Can Assign

Hard

You have n tasks and m workers. Each task has a strength requirement stored in a 0-indexed integer array tasks, with the ith task requiring tasks[i] strength to complete. The strength of each worker is stored in a 0-indexed integer array workers, with the jth worker having workers[j] strength. Each worker can only be assigned to a single task and must have a strength greater than or equal to the task’s strength requirement (i.e., workers[j] >= tasks[i]).

Additionally, you have pills magical pills that will increase a worker’s strength by strength. You can decide which workers receive the magical pills, however, you may only give each worker at most one magical pill.

Given the 0-indexed integer arrays tasks and workers and the integers pills and strength, return the maximum number of tasks that can be completed.

Example 1:

Input: tasks = [3,2,1], workers = [0,3,3], pills = 1, strength = 1

Output: 3

Explanation:

We can assign the magical pill and tasks as follows:

• Give the magical pill to worker 0.

• Assign worker 0 to task 2 (0 + 1 >= 1)

• Assign worker 1 to task 1 (3 >= 2)

• Assign worker 2 to task 0 (3 >= 3)

Example 2:

Input: tasks = [5,4], workers = [0,0,0], pills = 1, strength = 5

Output: 1

Explanation:

We can assign the magical pill and tasks as follows:

• Give the magical pill to worker 0.

• Assign worker 0 to task 0 (0 + 5 >= 5)

Example 3:

Input: tasks = [10,15,30], workers = [0,10,10,10,10], pills = 3, strength = 10

Output: 2

Explanation:

We can assign the magical pills and tasks as follows:

• Give the magical pill to worker 0 and worker 1.

• Assign worker 0 to task 0 (0 + 10 >= 10)

• Assign worker 1 to task 1 (10 + 10 >= 15)

The last pill is not given because it will not make any worker strong enough for the last task.

Constraints:

• n == tasks.length
• m == workers.length
• 1 <= n, m <= 5 * 104
• 0 <= pills <= m
• 0 <= tasks[i], workers[j], strength <= 109

Solution

import java.util.Deque
import java.util.LinkedList

@Suppress("NAME_SHADOWING")
class Solution {
    fun maxTaskAssign(tasks: IntArray, workers: IntArray, pills: Int, strength: Int): Int {
        var left = 0
        var right = tasks.size.coerceAtMost(workers.size)
        tasks.sort()
        workers.sort()
        while (left + 1 < right) {
            val mid = left + (right - left) / 2
            if (canAssign(mid, tasks, workers, pills, strength)) {
                left = mid
            } else {
                right = mid
            }
        }
        return if (canAssign(right, tasks, workers, pills, strength)) {
            right
        } else {
            left
        }
    }

    private fun canAssign(count: Int, tasks: IntArray, workers: IntArray, pills: Int, strength: Int): Boolean {
        var pills = pills
        val dq: Deque<Int> = LinkedList()
        var ind = workers.size - 1
        for (i in count - 1 downTo 0) {
            while (ind >= workers.size - count && workers[ind] + strength >= tasks[i]) {
                dq.offerLast(workers[ind])
                ind--
            }
            if (dq.isEmpty()) {
                return false
            }
            if (dq.peekFirst() >= tasks[i]) {
                dq.pollFirst()
            } else {
                dq.pollLast()
                pills--
                if (pills < 0) {
                    return false
                }
            }
        }
        return true
    }
}

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