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Author: jimit105

2180-maximum-number-of-tasks-you-can-assign/README.md

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+<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>

2180-maximum-number-of-tasks-you-can-assign/solution.py

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+from sortedcontainers import SortedList
+
+class Solution:
+    def maxTaskAssign(self, tasks: List[int], workers: List[int], pills: int, strength: int) -> int:
+        n, m = len(tasks), len(workers)
+        tasks.sort()
+        workers.sort()
+
+        def check(mid):
+            p = pills
+            ws = SortedList(workers[m - mid :])
+
+            # Enumerate each task from largest to smallest
+            for i in range(mid - 1, -1, -1):
+                if ws[-1] >= tasks[i]:
+                    ws.pop()
+                else:
+                    if p == 0:
+                        return False
+                    rep = ws.bisect_left(tasks[i] - strength)
+                    if rep == len(ws):
+                        return False
+                    p -= 1
+                    ws.pop(rep)
+            return True
+
+        left, right, ans = 1, min(m, n), 0
+        while left <= right:
+            mid = (left + right) // 2
+            if check(mid):
+                ans = mid
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        return ans
+
+