<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 n;
    int getNextIndex(vector<vector<int>> &temp,int left,int currJobEnd){
        int right = n-1;
        int result = n+1;
        while(left<=right){
            int mid = left + (right-left)/2;
            if(temp[mid][0]>=currJobEnd){
                result = mid;
                right = mid-1;
            }
            else{
                left = mid+1;
            }
        }
        return result;
    }
    int solve(vector<vector<int>> &temp,int index,vector<int> &dp){
        if(index>=n){
            return 0;
        }
        if(dp[index]!=-1){
            return dp[index];
        }

        int next = getNextIndex(temp,index+1,temp[index][1]);

        int taken = temp[index][2] + solve(temp,next,dp);
        int notTaken = 0 + solve(temp,index+1,dp);

        return dp[index] = max(taken,notTaken); 
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        n = startTime.size();
        vector<vector<int>> temp(n,vector<int>(3,0));
         for(int i = 0; i<n; i++) {
            temp[i][0] = startTime[i];
            temp[i][1] = endTime[i];
            temp[i][2] = profit[i];
        }
        sort(temp.begin(),temp.end());
        vector<int> dp(n+1,-1);
        return solve(temp,0,dp);
    }
};
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