<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:
   int next_index(vector<vector<int>>&arr , int index){
    int ans = arr.size()+1 ; 
    int low = index+1 ; 
    int high = arr.size()-1 ; 
    while(low<=high){
        int mid = (low+high)/2 ; 
        if(arr[index][1]<=arr[mid][0]){
            ans = mid; 
            high = mid-1 ; 
        }
        else {
            low = mid+1 ; 
        }
    }
    return ans ; 
   }
    int solve(vector<vector<int>>&arr , int index ,vector<int>&dp){
        if(index>=arr.size()){
            return 0 ; 
        }
        if(dp[index]!=-1){
            return dp[index] ;
        }
        int res = next_index(arr , index) ; 
        int take = arr[index][2]+solve(arr , res , dp) ; 
        int not_take = solve(arr , index+1 , dp) ; 
        return dp[index]= max(take , not_take) ; 
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n =startTime.size() ; 
        vector<int>dp(50001 , -1) ; 
        vector<vector<int>>arr(n , vector<int>(3,0)) ; 
        for(int i = 0 ; i<n ; i++){
            arr[i][0]=startTime[i] ; 
            arr[i][1]= endTime[i] ; 
            arr[i][2]= profit[i] ; 
        }
        sort(arr.begin() , arr.end()) ; 
        return solve(arr , 0 , dp) ; 
    }
};
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