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1
two sorted array A, B with length n,m (n <= m), and k which (k >= log n) is given.
with log (nm) we can find k-th smallest number in union of these two array.
I have a solution In problem 2 here, but my challenge is why two given condition "(n <= m)" and "k >= log n" does not affect this algorithm?
First assumption: n <= m is an assumption "without loss of generality". If n >= m, then just swap A and B in your head. They included this assumption even though it was not needed, because they felt it was "free" to make this assumption.
Second assumption: A trivial algorithm to find the kth smallest element is to iterate over A and B simultaneously, advancing in the array that has the smaller element of the two. This would be exactly like running the "Merge" function from mergesort, but stopping once you've merged the first k elements. The complexity would be O(k). They wanted you to find a more complex algorithm, so they "ruled out" this algorithm by stating that k >= log(n), which implies that complexity O(k) is never better than O(log(n)). Technically, if they wanted to thoroughly rule this algorithm out, they should also have stated that k <= n + m - log(n), otherwise you could run the "merge" function from the end: merge the n+m-k largest elements, then return the n+m-kth largest element, which is the same as the kth smallest element.
