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2071. Maximum Number of Tasks You Can Assign

Description

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

Solutions

Solution 1

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class Solution:
    def maxTaskAssign(
        self, tasks: List[int], workers: List[int], pills: int, strength: int
    ) -> int:
        def check(x):
            i = 0
            q = deque()
            p = pills
            for j in range(m - x, m):
                while i < x and tasks[i] <= workers[j] + strength:
                    q.append(tasks[i])
                    i += 1
                if not q:
                    return False
                if q[0] <= workers[j]:
                    q.popleft()
                elif p == 0:
                    return False
                else:
                    p -= 1
                    q.pop()
            return True

        n, m = len(tasks), len(workers)
        tasks.sort()
        workers.sort()
        left, right = 0, min(n, m)
        while left < right:
            mid = (left + right + 1) >> 1
            if check(mid):
                left = mid
            else:
                right = mid - 1
        return left
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class Solution {
    private int[] tasks;
    private int[] workers;
    private int strength;
    private int pills;
    private int m;
    private int n;

    public int maxTaskAssign(int[] tasks, int[] workers, int pills, int strength) {
        Arrays.sort(tasks);
        Arrays.sort(workers);
        this.tasks = tasks;
        this.workers = workers;
        this.strength = strength;
        this.pills = pills;
        n = tasks.length;
        m = workers.length;
        int left = 0, right = Math.min(m, n);
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (check(mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private boolean check(int x) {
        int i = 0;
        Deque<Integer> q = new ArrayDeque<>();
        int p = pills;
        for (int j = m - x; j < m; ++j) {
            while (i < x && tasks[i] <= workers[j] + strength) {
                q.offer(tasks[i++]);
            }
            if (q.isEmpty()) {
                return false;
            }
            if (q.peekFirst() <= workers[j]) {
                q.pollFirst();
            } else if (p == 0) {
                return false;
            } else {
                --p;
                q.pollLast();
            }
        }
        return true;
    }
}
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class Solution {
public:
    int maxTaskAssign(vector<int>& tasks, vector<int>& workers, int pills, int strength) {
        sort(tasks.begin(), tasks.end());
        sort(workers.begin(), workers.end());
        int n = tasks.size(), m = workers.size();
        int left = 0, right = min(m, n);
        auto check = [&](int x) {
            int p = pills;
            deque<int> q;
            int i = 0;
            for (int j = m - x; j < m; ++j) {
                while (i < x && tasks[i] <= workers[j] + strength) {
                    q.push_back(tasks[i++]);
                }
                if (q.empty()) {
                    return false;
                }
                if (q.front() <= workers[j]) {
                    q.pop_front();
                } else if (p == 0) {
                    return false;
                } else {
                    --p;
                    q.pop_back();
                }
            }
            return true;
        };
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (check(mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
};
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func maxTaskAssign(tasks []int, workers []int, pills int, strength int) int {
    sort.Ints(tasks)
    sort.Ints(workers)
    n, m := len(tasks), len(workers)
    left, right := 0, min(m, n)
    check := func(x int) bool {
        p := pills
        q := []int{}
        i := 0
        for j := m - x; j < m; j++ {
            for i < x && tasks[i] <= workers[j]+strength {
                q = append(q, tasks[i])
                i++
            }
            if len(q) == 0 {
                return false
            }
            if q[0] <= workers[j] {
                q = q[1:]
            } else if p == 0 {
                return false
            } else {
                p--
                q = q[:len(q)-1]
            }
        }
        return true
    }
    for left < right {
        mid := (left + right + 1) >> 1
        if check(mid) {
            left = mid
        } else {
            right = mid - 1
        }
    }
    return left
}

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