## first how we apply a greedy approach??

Here if look closely then we look at the arrival time of each meeting

1 if we see that if there exist a departure time which is smaller than the next arrival time then there is no need to allocate a separate room for it.

2 Based on this i have given a token +1 for entery and -1 for exit and sorted that array in order to know does first condition exist or not

## here is the code

```
ans=0
n=len(A)
k=[]
cur_ans=0
for i in range(n):
k.append([A[i][0],1])
k.append([A[i][1],-1])
k.sort()
for i in range(2*n):
ans+=k[i][1]
cur_ans=max(cur_ans,ans)
return cur_ans
```