LeetCode每日一题(2071. Maximum Number of Tasks You Can Assign)

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 * 10^4
• 0 <= pills <= m
• 0 <= tasks[i], workers[j], strength <= 10^9

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假设 tasks 和 workers 都是正序排列的， 且我们最多可以完成 k 个任务， 那我们一定完成的是 tasks[…k], 而这 k 个工人一定是 workers[workers.len() - k…]

而如果给出 k, 我们如何检测 tasks[…k]可以分配给 workers[workers.len() - k…]呢？

我们首先取出 tasks 中最大的， 即 tasks[k-1], 然后正序查找 workers 中是否有能不吃药就可以完成的， 如果有， 则将 worker 移除， 如果没有则正序查找 workers 看是否有能吃药完成该任务的， 如果有则将 worker 移除。如果以上两种方式都找不到， 则认为无法完成 k 个任务的分配。

整体我们用二分法来进行搜索， 最终确定 k

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impl Solution {
    fn can_complete(tasks: &[i32], workers: &[i32], mut pills: i32, strength: i32) -> bool {
        let mut workers = workers.to_vec();
        for &t in tasks.into_iter().rev() {
            let pos = match workers.binary_search(&t) {
                Ok(i) => i,
                Err(i) => i,
            };
            if pos < workers.len() {
                workers.remove(pos);
                continue;
            }
            if pills > 0 {
                pills -= 1;
                let pos = match workers.binary_search(&(t - strength)) {
                    Ok(i) => i,
                    Err(i) => i,
                };
                if pos < workers.len() {
                    workers.remove(pos);
                    continue;
                }
            }
            return false;
        }
        true
    }
    pub fn max_task_assign(mut tasks: Vec<i32>, mut workers: Vec<i32>, pills: i32, strength: i32) -> i32 {
        tasks.sort();
        workers.sort();
        let mut l = 0;
        let mut r = tasks.len().min(workers.len());
        while l < r {
            let m = (l + r) / 2;
            if Solution::can_complete(&tasks[..=m], &workers[workers.len() - m - 1..], pills, strength) {
                l = m + 1;
                continue;
            }
            r = m;
        }
        l as i32
    }
}