<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 Node:
      
      def __init__(self,s,e,v):
         self.s=s
         self.e=e
         self.v=v

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        
        dp={}
        node_list=[]
        for s,e,v in zip(startTime,endTime,profit):
             node_list.append(Node(s,e,v))
        node_list.sort(key=lambda x: x.e )
        
        
        def search(i):
        
            l,r=0,i-1

            while l<=r:
                    mid=(l+r)//2

                    if node_list[mid].e<=node_list[i].s:
                        
                        if node_list[mid+1].e<=node_list[i].s:
                             l=mid+1
                        else:
                             return mid
                    else:
                         r=mid-1

            return -1     



        dp[-1]=0
        dp[0]=node_list[0].v 
        
        ans=dp[0]
        for i in range(1,len(node_list)):
            j=search(i)
            dp[i]=max(dp[i-1],node_list[i].v+dp[j if j!=-1 else -1])
            ans=max(ans,dp[i])
        return ans               
                      

            


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