<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:
    //This is not a greedy problem
    //As you can see, there is a tradeoff between start time and endtime and profit
    //so, it's a dynamic programming question
    //How to do dp ?
    //sort the intervals based on time
    //dp[i] -> best possible profit you can make till ith interval
    //not pick case obvious -> best possible thing you can do till i-1
    //pick case -> if you are picking i -> then you may not pick i-1
    //You need to "search" for it, do a Binary Search

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();

        //For Sorting
        vector<int> indices(n);
        for(int i=0;i<n;i++){
            indices[i] = i;
        }    

        sort(indices.begin(),indices.end(),[&endTime,&startTime,&profit](int a,int b){
            if(endTime[a] != endTime[b])
                return endTime[a] < endTime[b]; //ascending order
            
            if(startTime[a] != startTime[b])
                return startTime[a] < startTime[b];

            return profit[a] > profit[b];
        });

        //Dynamic Programing
        vector<int> dp(n);
        dp[0] = profit[indices[0]]; //Base Case

        for(int i=1;i<n;i++){
            dp[i] = dp[i-1]; //not pick case

            //for pick case binary search
            int l = 0,r = i-1,ans = -1;
            while(l <= r){
                int m = l + (r-l)/2;
                if(endTime[indices[m]] > startTime[indices[i]]){
                    r = m-1;
                }
                else if(endTime[indices[m]] <= startTime[indices[i]]){
                    ans = m;
                    l = m+1;
                }
            }

            if(ans != -1){
                dp[i] = max(dp[i],dp[ans]+profit[indices[i]]);
            }
            else{
                dp[i] = max(dp[i],profit[indices[i]]);
            }
        }

        return dp[n-1];
    }
};
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