<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 func(int ind, int last, vector<vector<int>> ve,  vector<vector<int>>&dp) 
//     {
//         if (ind >= ve.size()) {
//             return 0;
//         }
//         if(last!=-1&&dp[ind][last]!=-1){
//             return dp[ind][last];
//         }

//         int pick = 0;
//         int np = 0;

//         if (last == -1 || ve[last][1] <= ve[ind][0]) 
//         {
//             pick = ve[ind][2] + func(ind + 1, ind, ve,dp);
//         }
      
//         np = func(ind + 1, last, ve,dp);
       
//        if(last!=-1){
//             dp[ind][last]=max(np,pick);
//         }

//         return max(np, pick);
//     }

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

//         vector<vector<int>> ve;
//         for (int i = 0; i < startTime.size(); i++) {
//             ve.push_back({startTime[i], endTime[i], profit[i]});
//         }
//         sort(ve.begin(), ve.end());
//         int ind = 0;
//         int last = -1;
//         vector<vector<int>>dp(ve.size(),vector<int>(ve.size(),-1));
//         return func(ind, last, ve,dp);
//     }
// };

class Solution {
public:
    int memo[50001];
    int n;
    //find the first job jiska starting point >= currentJob ka end point
    int getNextIndex(vector<vector<int>>& array, int l, int currentJobEnd) {
        int r = n-1;
        
        int result = n+1;
        
        while(l <= r) {
            int mid = l + (r-l)/2;
            
            if(array[mid][0] >= currentJobEnd) { //we can take this task
                result = mid;
                r = mid-1;
            } else {
                l = mid+1;
            }
        }
        
        return result;
    }
    
    int solve(vector<vector<int>>& array, int i) {
        if(i >= n)
            return 0;
        
        if(memo[i] != -1)
            return memo[i];
        
        int next  = getNextIndex(array, i+1, array[i][1]);
        
        int taken = array[i][2] + solve(array, next);
        int notTaken = solve(array, i+1);
        
        return memo[i] = max(taken, notTaken);
    }
    
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        n = startTime.size();
        
        memset(memo, -1, sizeof(memo));
        
        vector<vector<int>> array(n, vector<int>(3, 0)); //{s, e, p}
        
        for(int i = 0; i<n; i++) {
            array[i][0] = startTime[i];
            array[i][1] = endTime[i];
            array[i][2] = profit[i];
        }
        
       
        //sort kardo according to sart time
        sort(begin(array), end(array));
        
        return solve(array, 0);
    }
};
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