Maximum Number of Tasks You Can Assign
<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>
Greedy; Queue; Array; Two Pointers; Binary Search; Sorting; Monotonic Queue
class Solution {
public:
    int maxTaskAssign(vector<int>& tasks, vector<int>& workers, int pills, int strength) {
        sort(tasks.begin(), tasks.end());
        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) {
                    q.push_back(tasks[i++]);
                }
                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;
    }
};