- What is the Space complexity of quick sort algorithm? how do find it?
Solution: Quicksort has a space complexity of O(logn), even in the worst case, when it is carefully implemented such that
* in-place partitioning is used. This requires O(1).
* After partitioning, the partition with the fewest elements is (recursively) sorted first, requiring at most O(logn) space. Then the other partition is sorted using tail-recursion or iteration.
The version of quicksort with in-place partitioning uses only constant additional space before making any recursive call. However, if it has made O(logn) nested recursive calls, it needs to store a constant amount of information from each of them. Since the best case makes at most O(logn) nested recursive calls, it uses O(logn) space. The worst case makes O(n) nested recursive calls, and so needs O(n) space.However, if we consider sorting arbitrarily large lists, we have to keep in mind that our variables like left and right can no longer be considered to occupy constant space; it takes O(logn) bits to index into a list of n items. Because we have variables like this in every stack frame, in reality quicksort requires O(log2n) bits of space in the best and average case and O(nlogn) space in the worst case. This isn’t too terrible, though, since if the list contains mostly distinct elements, the list itself will also occupy O(nlogn) bits of space.
- What are dangling pointers?
Solution: A dangling pointer is a pointer to storage that is no longer allocated. Dangling pointers are nasty bugs because they seldom crash the program until long after they have been created, which makes them hard to find. Programs that create dangling pointers often appear to work on small inputs, but are likely to fail on large or complex inputs.
- Given that you can take one step or two steps forward from a given step. So find the total number of ways of reaching Nth step.
Solution:The simple recurrence relation governing this problem is f(N)=f(N-1) +f(N-2)(why?),which is a fibonacci sequence.
Nth state can be arrived directly by taking 2 step movement from N-2 or 1 step from N-1.Remember N-2 -> N-1 -> N is not a direct path from N-2th state to Nth state.Hence the no of solutions is no of ways to reach N-2th step and then directly taking a 2 jump step to N + no of ways to reach N-1th step and then taking 1 step advance. - You are given biased coin. Find unbiased decision out of it?
Solution:Throw the biased coin twice.Classify it as true for HT and false for TH.Both of these occur with probability=p*(1-p),hence unbiased. Ignore the other 2 events namely HH and TT.
- On a empty chessboard, a horse starts from a point( say location x,y) and it starts moving randomly, but once it moves out of board, it cant come inside. So what is the total probability that it stays within the board after N steps.
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