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
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
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
}
}