<p>You have <code>n</code> jobs and <code>m</code> workers. You are given three arrays: <code>difficulty</code>, <code>profit</code>, and <code>worker</code> where:</p>

<ul>
	<li><code>difficulty[i]</code> and <code>profit[i]</code> are the difficulty and the profit of the <code>i<sup>th</sup></code> job, and</li>
	<li><code>worker[j]</code> is the ability of <code>j<sup>th</sup></code> worker (i.e., the <code>j<sup>th</sup></code> worker can only complete a job with difficulty at most <code>worker[j]</code>).</li>
</ul>

<p>Every worker can be assigned <strong>at most one job</strong>, but one job can be <strong>completed multiple times</strong>.</p>

<ul>
	<li>For example, if three workers attempt the same job that pays <code>$1</code>, then the total profit will be <code>$3</code>. If a worker cannot complete any job, their profit is <code>$0</code>.</li>
</ul>

<p>Return the maximum profit we can achieve after assigning the workers to the jobs.</p>

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

<pre>
<strong>Input:</strong> difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
<strong>Output:</strong> 100
<strong>Explanation:</strong> Workers are assigned jobs of difficulty [4,4,6,6] and they get a profit of [20,20,30,30] separately.
</pre>

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

<pre>
<strong>Input:</strong> difficulty = [85,47,57], profit = [24,66,99], worker = [40,25,25]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>n == difficulty.length</code></li>
	<li><code>n == profit.length</code></li>
	<li><code>m == worker.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= difficulty[i], profit[i], worker[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def maxProfitAssignment(self, difficulty: List[int], profit: List[int], worker: List[int]) -> int:
        n = len(difficulty)
        profit_difficulty = [(profit[i], difficulty[i]) for i in range(n)]
        profit_difficulty.sort(reverse=True)

        worker.sort()

        res = 0
        for p, d in profit_difficulty:
            while worker and worker[-1] >= d:
                worker.pop()
                res += p
        return res
        
18800