<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 globalMaxProfit = 0;

        // Combine of binary search and max subarray max
        std::unordered_map<int, int> UMAP;
        for(int i = 0; i < difficulty.size(); ++i)
        {
            auto it = UMAP.find(difficulty.at(i));
            if(it != UMAP.end())
            {
                it->second = std::max(it->second, profit.at(i));
            }
            else UMAP[difficulty.at(i)] = profit.at(i);
        }

        std::sort(difficulty.begin(), difficulty.end());

        // Maximum subarray max
        std::vector<int> MSAM;
        MSAM.reserve(difficulty.size());
        int previousMSAM = UMAP[difficulty.front()];
        for(auto& d : difficulty)
        {
            previousMSAM = std::max(previousMSAM, UMAP[d]);
            MSAM.emplace_back(previousMSAM);
            std::cout << "MSAM: " << previousMSAM << ", difficulty = " << d << std::endl;
        }

        for(auto& w : worker)
        {
            int maxPossibleDifficulty = -1;
            int left = 0;
            int right = difficulty.size() - 1;
            while(left <= right)
            {
                int mid = left + (right - left) / 2;
                if(difficulty.at(mid) == w)
                {
                    maxPossibleDifficulty = mid;
                    break;
                }
                else if(difficulty.at(mid) > w) right = mid - 1;
                else
                {
                    maxPossibleDifficulty = mid;
                    left = mid + 1;
                }
            }

            int test = maxPossibleDifficulty == -1 ? 0 : MSAM.at(maxPossibleDifficulty);
            std::cout << "MSAM.at(maxPossibleDifficulty) = " << test << std::endl;
            globalMaxProfit += maxPossibleDifficulty == -1 ? 0 : MSAM.at(maxPossibleDifficulty);
        }

        return globalMaxProfit;
    }
};


// 35 47 52 68 86
// |  |  |  |  |
// 17 81 1  67 3
// 17 81 81 81 81
45584