I used a dfs for finding no. of connected component. If this return >1 then answer would be 0. Can you please explain this?
@ay2306 Actually in this case, my code could pass but i have found another case something like u mentioned above. It was my fault, i forgot that cases and thats why got partial marks.
@anon87574747 Yes, my code failed for tc2 and 9. Actually i had some logical mistake so got partial marks.
My code failed for:
ggdgg
ggggg
ggdgg
@aman_2k18 It means i have to make graph disjoint i.e. if i start with any ‘g’ index then not able to visit each ‘g’ index. So, i am using dfs starting with any ‘g’ index and check whether i am able to visit each ‘g’ index or not. If not, then it proves that this graph contains more than 1 ‘g’ connected component. Hence answer would be 0.
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I think almost all of us failed for the test case when g\leq 2 and they are connected. In that case, we have to remove both g's. Because removing only one would make the other g connected with itself. But as per our logic, we were removing only one of them. So basically, when g\leq 2, then print g as the answer.
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even i tried doing the same thing but got WA on all testcases. Later used binary search with prefix sum and it worked. though the logic is still the same.
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Same question, is it restricted to only a few colleges? I hope someone replies
Makes sense ! Idk why I used algorithm for finding articulation point.
What about
d d d
d g d
d d d
Case ?
Ans should be 1 right ?
Had you missed this case or am I missing anything ?
Edit : Just read it fails with many more cases.
I did this.
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Did you get full score on the question?
how exactly ???
Atleast I know, our approach is correct. 
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Ohh, I didn’t know.
I think at least the test files were correct.
which contest is this ques of?
That’s secret 
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nice one
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