<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 {
public:

    static bool sort_key(vector<int> &a, vector<int> &b) {
        return a[0] < b[0];
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {

        vector<vector<int>> jobs;

        for(int i=0; i<profit.size(); i++) {
            jobs.push_back( vector<int>{startTime[i], endTime[i], profit[i]} );
        }

        sort(jobs.begin(), jobs.end(), sort_key);
        map<int,int> memo;

        return max_profit(jobs, 0, 0, memo);
    }

    int bin_search(vector<vector<int>> &jobs, int end_time, int low, int high) {

        int last_index = high;

        while(low <= high) {
            int mid = low + (high - low) / 2;

            if(jobs[mid][0] >= end_time) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        if(low == last_index+1) return -1;

        return low;
    }

    int max_profit(vector<vector<int>> &jobs, int current_index, int prev_end_time, map<int,int> &memo) {

        if(current_index >= jobs.size()) return 0;

        if(memo.find(current_index) != memo.end()) {
            return memo[current_index];
        }

        int current_start_time = jobs[current_index][0];
        int profit_if_job_included = 0;
        if(prev_end_time <= current_start_time) {
            // Can include the job
            int profit_from_this_job = jobs[current_index][2];
            int current_end_time = jobs[current_index][1];

            int next_doable_job_index = bin_search(jobs, current_end_time, current_index+1, jobs.size()-1);

            if(next_doable_job_index == -1) {
                profit_if_job_included = profit_from_this_job;
            } else {
                int next_job_sub_tree_profit = max_profit(jobs, next_doable_job_index, current_end_time, memo);
                profit_if_job_included = next_job_sub_tree_profit + profit_from_this_job;
            }
            
        }

        // Current Job not included

        int profit_if_job_not_included = max_profit(jobs, current_index+1, prev_end_time, memo);

        memo[current_index] = max(profit_if_job_included, profit_if_job_not_included);


        return max(profit_if_job_included, profit_if_job_not_included);
    }
};
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