Hey Ankur, What is the order of time complexity we are looking for in this case. The option which Dave suggested can give us random node by traversing that many number of nodes from the head. That will be O(n).
This can be further reduced to n/2 if we use two pointers, both of which will traverse two nodes at a time: 1. one pointing to first node (lets call it odd pointer) 2. other pointing to second node (lets call it even pointer) So, depending on the value returned by random number generator(even or odd), we can decide which pointer to pick. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/N-5i9YH4AkYJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
