Finding kth smallest number from n sorted arrays

This doesn't generalize the links, but does solve the problem:

1. Go through all the arrays and if any have length > k, truncate to length k (this is silly, but we'll mess with k later, so do it anyway)
2. Identify the largest remaining array A. If more than one, pick one.
3. Pick the middle element M of the largest array A.
4. Use a binary search on the remaining arrays to find the same element (or the largest element <= M).
5. Based on the indexes of the various elements, calculate the total number of elements <= M and > M. This should give you two numbers: L, the number <= M and G, the number > M
6. If k < L, truncate all the arrays at the split points you've found and iterate on the smaller arrays (use the bottom halves).
7. If k > L, truncate all the arrays at the split points you've found and iterate on the smaller arrays (use the top halves, and search for element (k-L).

When you get to the point where you only have one element per array (or 0), make a new array of size n with those data, sort, and pick the kth element.

Because you're always guaranteed to remove at least half of one array, in N iterations, you'll get rid of half the elements. That means there are N log k iterations. Each iteration is of order N log k (due to the binary searches), so the whole thing is N^2 (log k)^2 That's all, of course, worst case, based on the assumption that you only get rid of half of the largest array, not of the other arrays. In practice, I imagine the typical performance would be quite a bit better than the worst case.

It can not be done in less than O(n) time. Proof Sketch If it did, it would have to completely not look at at least one array. Obviously, one array can arbitrarily change the value of the kth element.

I have a relatively simple O(n*log(n)*log(m)) where m is the length of the longest array. I'm sure it is possible to be slightly faster, but not a lot faster.

Consider the simple case where you have n arrays each of length 1. Obviously, this is isomorphic to finding the kth element in an unsorted list of length n. It is possible to find this in O(n), see Median of Medians algorithm, originally by Blum, Floyd, Pratt, Rivest and Tarjan, and no (asymptotically) faster algorithms are possible.

Now the problem is how to expand this to longer sorted arrays. Here is the algorithm: Find the median of each array. Sort the list of tuples (median,length of array/2) and sort it by median. Walk through keeping a sum of the lengths, until you reach a sum greater than k. You now have a pair of medians, such that you know the kth element is between them. Now for each median, we know if the kth is greater or less than it, so we can throw away half of each array. Repeat. Once the arrays are all one element long (or less), we use the selection algorithm.

Implementing this will reveal additional complexities and edge conditions, but nothing that increases the asymptotic complexity. Each step

1. Finds the medians or the arrays, O(1) each, so O(n) total
2. Sorts the medians O(n log n)
3. Walks through the sorted list O(n)
4. Slices the arrays O(1) each so, O(n) total

that is O(n) + O(n log n) + O(n) + O(n) = O(n log n). And, we must perform this untill the longest array is length 1, which will take log m steps for a total of O(n*log(n)*log(m))

You ask if this can be generalized to the case of unsorted arrays. Sadly, the answer is no. Consider the case where we only have one array, then the best algorithm will have to compare at least once with each element for a total of O(m). If there were a faster solution for n unsorted arrays, then we could implement selection by splitting our single array into n parts. Since we just proved selection is O(m), we are stuck.

You could look at my recent answer on the related question here. The same idea can be generalized to multiple arrays instead of 2. In each iteration you could reject the second half of the array with the largest middle element if k is less than sum of mid indexes of all arrays. Alternately, you could reject the first half of the array with the smallest middle element if k is greater than sum of mid indexes of all arrays, adjust k. Keep doing this until you have all but one array reduced to 0 in length. The answer is kth element of the last array which wasn't stripped to 0 elements.

Run-time analysis:

You get rid of half of one array in each iteration. But to determine which array is going to be reduced, you spend time linear to the number of arrays. Assume each array is of the same length, the run time is going to be cclog(n), where c is the number of arrays and n is the length of each array.