If we can have additional O(n) space than all operations wil take O(1) ..additional MinStack implementation..
But if we are not allowed O(n) space then at the cost of GetMinimum to become O(1) we have to sacrifice for push and pop to be O(n)..{this is not mentioned in option} So most efficient will be O(1) time for each -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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.