<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 findNext(  vector<vector<int>> &vec, int endTime){
        int lo = 0;
        int hi=N-1;
        int ans=N;
        while(lo<=hi){
            int mid =lo+(hi-lo)/2;
           
            if (vec[mid][0]>=endTime){
                hi= mid-1;
            }else{
                ans= mid;
                lo=mid+1;
            }
        }
        return ans;
    }
    map<int,int>dp;
    int N;
    int dfs(vector<vector<int>> &input, int idx){
        if(idx>=N){
            return 0;
        }
        if(dp.find(idx)!=dp.end()){
            return dp[idx];
        }
       // int nextEndIdx = findNext(input, input[idx][1]);
        int lo = 0, hi = N - 1, newJobTime = N + 1;
        while(lo <= hi){
            int mid = (lo + hi) >> 1;
            if(input[mid][0] >= input[idx][1]) newJobTime = mid, hi = mid - 1;
            else lo = mid + 1;
        }
        int skp=dfs(input, idx+1);
        int take=dfs(input,newJobTime)+input[idx][2];
        return dp[idx]=max(skp,take);
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<vector<int>> vec;
        for(int i=0;i<startTime.size();i++)
        vec.push_back({startTime[i],endTime[i],profit[i]});

      //  vector<int>dp(vec.size()+2,0);
        sort(vec.begin(),vec.end());
      dp.clear();
      N=vec.size();
        return dfs(vec,0);   
    }
};
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