<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>

#include <iostream>
#include <iomanip>

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cassert>
#include <ctime>

#include <algorithm>
#include <utility>

#include <vector>
#include <queue>
#include <set>
#include <string>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define lb long double
#define feq(a, b) (fabs(a - b) < eps)
#define f0r(n, h) for (int h = 0; h < n; ++h)
#define ln "\n"

const int inf = 1e9;
const ll llinf = 1e15;
const int MOD = 1e9 + 7;
const ll LLMOD = (1 << (23)) * 119 + 1;
const int MX = 2e5 + 10;
const lb eps = 1e-9;
class Solution {
public:
    int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
           cin.tie(0)->sync_with_stdio(0);
   cout.precision(18);
        sort(worker.begin(),worker.end());
        int n=difficulty.size();
        int m=worker.size();
        pii arr[n];
        f0r(n,i){
            arr[i]=mp(profit[i],difficulty[i]);
        }
        sort(arr,arr+n);
        int wind=m-1;
        int jind=n-1;
        int ans=0;
        while(jind>=0){
            while(wind>-1&&worker[wind]>=arr[jind].ss){
                ans+=arr[jind].ff;
                wind--;
            }
            jind--;
        }
        return ans;
    }
};
18853