<p>We have <code>n</code> jobs, where every job is scheduled to be done from <code>startTime[i]</code> to <code>endTime[i]</code>, obtaining a profit of <code>profit[i]</code>.</p>

<p>You&#39;re given the <code>startTime</code>, <code>endTime</code> and <code>profit</code> arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.</p>

<p>If you choose a job that ends at time <code>X</code> you will be able to start another job that starts at time <code>X</code>.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png" style="width: 380px; height: 154px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
<strong>Output:</strong> 120
<strong>Explanation:</strong> The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png" style="width: 600px; height: 112px;" /> </strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
<strong>Output:</strong> 150
<strong>Explanation:</strong> The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png" style="width: 400px; height: 112px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= startTime.length == endTime.length == profit.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= startTime[i] &lt; endTime[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= profit[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
private:
    int getMaxProfit(int startInd, vector<int>& startTime, unordered_map<int, vector<pair<int, int>>> &startInfo, int n) {
        if (startInd == n) return 0;

        // not take any startJob
        int notTake = 0 + getMaxProfit(startInd + 1, startTime, startInfo, n);

        // take all possible jobs starting at startTime[startInd]
        int take = INT_MIN;
        int startJob = startTime[startInd];
        for (auto it: startInfo[startJob]) {
            int endTime = it.first, profit = it.second;

            int nextStartInd = lower_bound(startTime.begin(), startTime.end(), endTime) - startTime.begin();
            int currProfit = profit + getMaxProfit(nextStartInd, startTime, startInfo, n);

            take = max(take, currProfit);
        }

        return max(take, notTake);
    }
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        map<int, int> times;
        unordered_map<int, vector<pair<int, int>>> jobs;
        for (auto i = 0; i < startTime.size(); ++i) {
            times[startTime[i]] = 0;
            jobs[startTime[i]].push_back({ endTime[i], profit[i] });
        }    
        int max_profit = 0;
        for (auto it = rbegin(times); it != rend(times); ++it) {
            for (auto job : jobs[it->first]) {
                auto it = times.lower_bound(job.first);
                max_profit = max(max_profit, (it == end(times) ? 0 : it->second) + job.second);
            }
            it->second = max_profit;
        }
        return max_profit;
    }  
};
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