<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>

 bool cmp(vector<int>&v1,vector<int>&v2){
    return v1[0] < v2[0];
}

class Solution {
public:
    int memo[50001];
    int n;
    int bSEARCH(vector<vector<int>>&nums,int low,int currentJobEnd){
        int high = nums.size()-1;
        int ans = n; // u can put any variable of your own choice
        while(low <= high){
            int mid = low + (high-low)/2;
            if(nums[mid][0] >= currentJobEnd){ // we can take this as a next index
                ans = mid;
                high = mid-1; // aur chhota diff mil skta hai isliye using binary search

            }
            else low = mid+1;
        }
        return ans;
    }

    int solve(vector<vector<int>>&nums,int i){
        // base case
        if(i >= n)return 0; // index goes out of bound so no profit will be get
        // main code
        //starting time must be more or equal to ending time to tart a new job
        // taken
        if(memo[i] != -1)return memo[i]; 
        int nextIdx = bSEARCH(nums,i+1,nums[i][1]); // passinf idx+1 and current job end value
        int taken = nums[i][2] +  solve(nums,nextIdx);
        // not taken
        int notTaken = 0 + solve(nums,i+1); // next index will be idx + 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>>nums(n,vector<int>(3,0)); // {{1,3,50},{2,4,10},{3,5,40}}.... merge all vector in ascending order
        for(int i=0;i<n;i++){
            nums[i][0] = startTime[i];
            nums[i][1] = endTime[i];
            nums[i][2] = profit[i];
        }

        // sort them in ascending order
        sort(nums.begin(),nums.end(),cmp);
       


        return solve(nums,0); // calling vy passing index and array(nums)
    }
};
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