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Auto comment: topic has been updated by Parvej (previous revision, new revision, compare).
Just try to backtracking/bruteforces. Each vertices during the visit will be marked (not go to that vertice again), else unmark it. The complexity will be $$$O(n!)$$$ time
SPyofgame I think time complexity will be O ((N^2) * (N!)) as for each path there will be N^2 computation right? Correct me if I am wrong. Thanks in advance.
I dont think we need $$$O(n^2)$$$ for each $$$O(n!)$$$ path since you can try to implement directly or something similar.
By the way, $$$O(n^2)$$$ is also considerably small compared to $$$O(n!)$$$. So unless the time limit is small, you can try to improve the complexity by changing implementation or even a bit of heuristic — which can provide better complexity and constant.