To find an element in a set or a map take the same time. It is
a) O(1)
b) O(N1/2)
c) O(log N)
d) O(N)
e) O(N2)
c) O(log N)
The set and map keep their data in a binary search tree that is kept balanced. I won’t go into the details, but is the tree is kept close enough to full that the number of elements necessary to be examined approximates the base 2 log of the number of elements. Draw some pictures and reflect a little. You will be convinced that this gives a bound of O(log N).
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