I was recently asked this question in an interview:
There are two arrays of size \'n\' each. One array has nuts, the other one has bolts. Each nut fits exactly one bolt
Take one nut N0
and compare it against all bolts. With the resulting information, we can split the bolts array into [bolts smaller than B0] + B0 + [bolts larger than B0]
. There is always a unique B0
that fits N0
based on the statement of the question.
Then take the next nut N1
and compare it against B0
. If the result is "tight", we search the smaller half as we did above with N0
. Otherwise, we do the same but with the larger half. Doing this will further split one of the two halves into 2.
Continue doing this until you've worked through all nuts. This is equivalent to quicksort. Average case is O(N logN), but there's the obvious worst case complexity of O(N^2) when the list is "sorted" already.