<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>

class Solution:
    def maxTaskAssign(self, tasks: List[int], workers: List[int], pills: int, strength: int) -> int:
        # workers sorted in reverse order, tasks sorted in normal order
        def can_assign(n: int) -> bool:
            """
            Check if it's possible to assign `n` tasks to the `n` strongest workers.
            """
            # Index for tasks
            task_idx = 0
            # Deque to temporarily hold tasks that can potentially be assigned
            task_temp = deque()
            # Track the number of pills available
            n_pills = pills

            # Traverse from the strongest available worker to the weakest in the subset of `n` workers
            for worker_idx in range(n - 1, -1, -1):
                # Add tasks that the current worker can do, even with the help of a pill
                while task_idx < n and tasks[task_idx] <= workers[worker_idx] + strength:
                    task_temp.append(tasks[task_idx])
                    task_idx += 1

                # If no tasks can be assigned, return False
                if len(task_temp) == 0:
                    return False

                # If the worker can do the weakest task without a pill
                if workers[worker_idx] >= task_temp[0]:
                    task_temp.popleft()  # Assign the task to the worker
                elif n_pills > 0:  # Otherwise, use a pill if available
                    task_temp.pop()  # Assign the hardest possible task with the pill
                    n_pills -= 1
                else:
                    return False  # If no task can be assigned, return False

            return True  # All tasks in the subset can be assigned

        # Sort tasks in increasing order and workers in decreasing order
        tasks.sort()
        workers.sort(reverse=True)

        # Initialize binary search boundaries
        left, right = 0, min(len(tasks), len(workers))
        res = 0
        
        # Binary search to find the maximum number of tasks that can be assigned
        while left <= right:
            mid = (left + right + 1) // 2
            if can_assign(mid):  # If it's possible to assign `m` tasks
                res = mid  # Update result
                left = mid + 1  # Try for more tasks
            else:
                right = mid - 1  # Try fewer tasks
        
        return res
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