<p>You have <code>n</code> tasks and <code>m</code> workers. Each task has a strength requirement stored in a <strong>0-indexed</strong> integer array <code>tasks</code>, with the <code>i<sup>th</sup></code> task requiring <code>tasks[i]</code> strength to complete. The strength of each worker is stored in a <strong>0-indexed</strong> integer array <code>workers</code>, with the <code>j<sup>th</sup></code> worker having <code>workers[j]</code> strength. Each worker can only be assigned to a <strong>single</strong> task and must have a strength <strong>greater than or equal</strong> to the task&#39;s strength requirement (i.e., <code>workers[j] &gt;= tasks[i]</code>).</p>

<p>Additionally, you have <code>pills</code> magical pills that will <strong>increase a worker&#39;s strength</strong> by <code>strength</code>. You can decide which workers receive the magical pills, however, you may only give each worker <strong>at most one</strong> magical pill.</p>

<p>Given the <strong>0-indexed </strong>integer arrays <code>tasks</code> and <code>workers</code> and the integers <code>pills</code> and <code>strength</code>, return <em>the <strong>maximum</strong> number of tasks that can be completed.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> tasks = [<u><strong>3</strong></u>,<u><strong>2</strong></u>,<u><strong>1</strong></u>], workers = [<u><strong>0</strong></u>,<u><strong>3</strong></u>,<u><strong>3</strong></u>], pills = 1, strength = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong>
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 &gt;= 1)
- Assign worker 1 to task 1 (3 &gt;= 2)
- Assign worker 2 to task 0 (3 &gt;= 3)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> tasks = [<u><strong>5</strong></u>,4], workers = [<u><strong>0</strong></u>,0,0], pills = 1, strength = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong>
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 &gt;= 5)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> tasks = [<u><strong>10</strong></u>,<u><strong>15</strong></u>,30], workers = [<u><strong>0</strong></u>,<u><strong>10</strong></u>,10,10,10], pills = 3, strength = 10
<strong>Output:</strong> 2
<strong>Explanation:</strong>
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 &gt;= 10)
- Assign worker 1 to task 1 (10 + 10 &gt;= 15)
The last pill is not given because it will not make any worker strong enough for the last task.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == tasks.length</code></li>
	<li><code>m == workers.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= pills &lt;= m</code></li>
	<li><code>0 &lt;= tasks[i], workers[j], strength &lt;= 10<sup>9</sup></code></li>
</ul>

from typing import List
from sortedcontainers import SortedList

class Solution:
    def maxTaskAssign(self, tasks: List[int], workers: List[int], pills: int, strength: int) -> int:
        tasks.sort()
        workers.sort()
        
        def canComplete(k):
            available_workers = SortedList(workers[-k:])
            p = pills
            
            for task in reversed(tasks[:k]):
                # Try to assign the strongest available worker who can complete the task without a pill
                idx = available_workers.bisect_left(task)
                
                if idx < len(available_workers):
                    available_workers.pop(idx)
                else:
                    # If no such worker is available, try using a pill
                    idx = available_workers.bisect_left(task - strength)
                    if p > 0 and idx < len(available_workers):
                        available_workers.pop(idx)
                        p -= 1
                    else:
                        return False
            return True
        
        low, high = 0, min(len(tasks), len(workers))
        while low < high:
            mid = (low + high + 1) // 2
            if canComplete(mid):
                low = mid
            else:
                high = mid - 1
        
        return low

# Example usage
solution = Solution()
print(solution.maxTaskAssign([3, 2, 1], [0, 3, 3], 1, 1))  # Output: 3
print(solution.maxTaskAssign([5, 4], [0, 0, 0], 1, 5))    # Output: 1
print(solution.maxTaskAssign([10, 15, 30], [0, 10, 10, 10, 10], 3, 10))  # Output: 2
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