<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 {
public:
    int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
        vector<pair<int,int>> vec(difficulty.size());
        for(int i=0; i<difficulty.size(); i++){
            vec[i] = {difficulty[i], profit[i]};
        }
        auto comp = [&](pair<int,int> a, pair<int,int> b){
            if(a.first < b.first)return true;
            return false;
        };
        sort(vec.begin(), vec.end(), comp);
        sort(worker.begin(), worker.end());
        int res = 0;
        int cov = 0;
        int maxProf = 0;
        for(int i=0; i<worker.size(); i++){
            int getMid = binarySearch(vec, worker[i]);
            for(int i=cov; i<=getMid; i++){
                maxProf = max(maxProf, vec[i].second);
            }
            cov = max(0,getMid);
            res += maxProf;
        }
        return res;
    }
    int binarySearch(vector<pair<int,int>>& vec,int target){
        int l = 0;
        int r = vec.size()-1;
        int res = -1;
        while(l<=r){
            int mid = (l+r)/2;
            if(vec[mid].first <= target){
                res = mid;
                l = mid + 1;
            }else{
                r = mid - 1;
            } 
        }
        return res;
    }
};
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