# 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

• class Solution {
private int[] workers;
private int strength;
private int pills;
private int m;
private int n;

Arrays.sort(workers);
this.workers = workers;
this.strength = strength;
this.pills = pills;
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) {
}
if (q.isEmpty()) {
return false;
}
if (q.peekFirst() <= workers[j]) {
q.pollFirst();
} else if (p == 0) {
return false;
} else {
--p;
q.pollLast();
}
}
return true;
}
}

• class Solution {
public:
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) {
}
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;
}
};

• class Solution:
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:
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

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


• func maxTaskAssign(tasks []int, workers []int, pills int, strength int) int {
sort.Ints(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 {
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
}