didn't understand, for scanning the adjacency list E, the complexity will O(n), hence the Overall complexity of the problem becomes O(n^2). Unless we don't make an adjacency list as an Array of length N, so our search in adjacency list become O(1).
i guess if u are using a hashmap adjacency list u can simply determine if a node is present or not in O(1) but if you are using a vector then I guess yr answer will be right since if u have to know if a node has a link to you ud have to traverse that vector in O(n) and doing this n times get n^2 complexity
Finally understood the algorithm completely! thank you sir.
yeah
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Practice problems on this topic?
thank u so much sir ,,,,,,its really helpfull for me
yeah
this is 100x better than my china prof
didn't understand, for scanning the adjacency list E, the complexity will O(n), hence the Overall
complexity of the problem becomes O(n^2). Unless we don't make an adjacency list as an Array of length N,
so our search in adjacency list become O(1).
i guess if u are using a hashmap adjacency list u can simply determine if a node is present or not in O(1) but if you are using a vector then I guess yr answer will be right since if u have to know if a node has a link to you ud have to traverse that vector in O(n) and doing this n times get n^2 complexity
Thank you sir.
thank you
Badia videos sirji
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