# Order Statistics

How do you find the kth biggest element in an array of size n?

A naive $O(n\log{n})$ method would be to sort the array and return the element arr[n-k]. But we are looking for something linear in the length of the array.

Now, this comes from one of the CSE548 lectures of Prof. Michael A. Bender.

Let us look at a Randomized Algorithm approach to this:  Randomly pick a pivot p

Partition the elements into the left sub-array with elements <= p and right sub-array with elements > p respectively.

If rank(p) = k, then return p

If rank(p) > k, then return the k th element from the left sub-array

If rank(p) < k, then return the k-rank(p) th element from the right sub-array

This definitely smells like $O(n\^2)$. But turns out, in the average case it is not.

If you assume that your pivots are all good, and there is a 1/2 probability of choosing a good pivot (your pivot has a rank between n/4 and 3n/4, inclusive), your recurrence relation is only as bad as this, $T(n) = T(3n/4) + O(n)$

Which is, $T(n) = \theta(n)$.

Surprisingly this can be done deterministically. An ‘All-Stars’ deterministic algorithm devised by Blum, Floyd, Pratt, Rivest, Tarjan, does this in $O(n)$. I haven’t read it yet though.

One of the new tools that Prof. Bender hands you in CSE548, is the notion of Randomized Algorithms, which work ‘with high probability’. I will probably go deeper in that field later.