this code is giving correct answer for sample testcase.
But not passing other test cases due to timelimit
plzz suggest something
Beautiful vertices : graph
This question is an interesting implementation of DFS. Perform DFS while maintaining a count of children of the vertices that you have visited.
Let u be a vertex and v be its child. Then, check if children[v]>children[u].This way, check for each vertex while traversing the graph in DFS.
NOTE:
Actually there are many trees in the input(a graph with no cycles and self loops is a tree). That means , you have to apply DFS for each of these trees.
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