<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) {
        int jobSize = difficulty.size();
        vector<vector<int>> difficultyAndProfit(jobSize);
        for (int i = 0; i < jobSize; i++) {
            int& difficultyElement = difficulty.at(i);
            int& profitElement = profit.at(i);
            vector<int> difficultyAndProfitElement = {difficultyElement, profitElement};
            difficultyAndProfit.at(i) = difficultyAndProfitElement;
        }

        sort(difficultyAndProfit.begin(), difficultyAndProfit.end(), [](vector<int>& a, vector<int>& b) {
            return a.at(0) < b.at(0);
        });
        for (int i = 1; i < jobSize; i++) {
            vector<int>& difficultyAndProfitPrevElement = difficultyAndProfit.at(i - 1);
            vector<int>& difficultyAndProfitCurElement = difficultyAndProfit.at(i);
            int prevProfit = difficultyAndProfitPrevElement.at(1);
            int curProfit = difficultyAndProfitCurElement.at(1);
            difficultyAndProfitCurElement.at(1) = max(prevProfit, curProfit);
        }

        int result = 0;
        for (auto& workerAbility : worker) {
            int workerMaxProfit = binarySearch(difficultyAndProfit, workerAbility);
            result += workerMaxProfit;
        }

        return result;
    }
    int binarySearch(vector<vector<int>>& difficultyAndProfit, int workerAbility) {
        int difficultyAndProfitSize = difficultyAndProfit.size();
        int result = 0;
        int left = 0;
        int right = difficultyAndProfitSize - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            vector<int>& difficultyAndProfitElement = difficultyAndProfit.at(mid);
            int& jobDifficulty = difficultyAndProfitElement.at(0);
            int& jobProfit = difficultyAndProfitElement.at(1);
            if (jobDifficulty <= workerAbility) {
                result = jobProfit;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return result;
    }
};
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