<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 getIndex(vector<pair<int,pair<int,int>>> &arr,int l,int target){
        int r = arr.size()-1;
        while(l <= r){
            int mid = l + (r-l)/2;
            if(arr[mid].second.first < target){
                l = mid+1;
            }
            else{
                r = mid-1;
            }
        }
        return l;
    }

    int solve(vector<pair<int,pair<int,int>>> &arr,int index){
        if(index >= arr.size()){
            return 0;
        }
        int next = getIndex(arr,index+1,arr[index].second.second);
        int inc = arr[index].first + solve(arr,next);
        int exc = solve(arr,index+1);
        return max(inc,exc);
       
    }

    int solveMem(vector<pair<int,pair<int,int>>> &arr,int index,vector<int> &dp){
        if(index >= arr.size()){
            return 0;
        }

        if(dp[index] != -1){
            return dp[index];
        }
        
        int next = getIndex(arr,index+1,arr[index].second.second);
        int inc = arr[index].first + solveMem(arr,next,dp);
        int exc = solveMem(arr,index+1,dp);
        return dp[index] = max(inc,exc);
       
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<pair<int,pair<int,int>>> arr(profit.size());
        for(int i=0; i<profit.size(); i++){
            arr[i].first = profit[i];
            arr[i].second.first = startTime[i];
            arr[i].second.second = endTime[i];
        }
        auto cmp = [](pair<int,pair<int,int>> &a,pair<int,pair<int,int>> &b){
            return a.second.first < b.second.first;
        };
        sort(arr.begin(),arr.end(),cmp);
        // return solve(arr,0);
        vector<int> dp(arr.size()+1,-1);
        return solveMem(arr,0,dp);
    }
};
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