Question

I have this question:

Given two sorted lists (stored in arrays) of size n, find an O(log n)
algorithm that computes the nth largest element in the union of the
two lists.

I can see there is probably a trick here as it requires the nth largest element and the arrays are also of size n, but I can’t figure out what it is. I was thinking that I could adapt counting sort, would that work?

Answer 1

Compare A[n/2] and B[n/2]. If equal, any of them is our result. Other stopping condition for this algorithm is when both arrays are of size 1 (either initially or after several recursion steps). In this case we just choose the largest of A[n/2] and B[n/2].

If A[n/2] < B[n/2], repeat this procedure recursively for second half of A[] and first half of B[].

If A[n/2] > B[n/2], repeat this procedure recursively for second half of B[] and first half of A[].

Since on each step the problem size is (in worst case) halved, we’ll get O(log n) algorithm.

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Always dividing array size by two to get the index works properly only if n is a power of two. More correct way of choosing indexes (for arbitrary n) would be using the same strategy for one array but choosing complementing index: j=n-i for other one.