Problem : Given an unsorted array, find the kth smallest element. The problem is called the Selection problem . It's been intensively studied and has a couple of very interesting algorithms that do the job. I'll be describing an algorithm called QuickSelect . The algorithm derives its name from QuickSort. You will probably recognise that most of the code directly borrows from QuickSort. The only difference being there is a single recursive call rather than 2 in QuickSort. The naive solution is obvious, simply sort the array `O(nlogn)` and return the kth element. Infact, you can partially sort it and use the Selection sort to get the solution in `O(nk)` An interesting side effect of finding the kth smallest element is you end up finding the k smallest elements. This also effectively gives you (n - k) largest elements in the array as well. These elements are not in any particular order though. The version I'm using uses a random pivot selection, this part of the al
Given an Array A[1..n] of distinct integers find out how many instances are there where A[i] > A[j] where i < j. Such instances can be called Inversions. Say, [100,20,10,30] the inversions are (100,20), (100, 10), (100,30), (20,10) and the count is of course 4 The question appears as an exercise in the Chapter 4 of Introduction to Algorithms . (3rd Edition) As per our tradition we'll start with the worst algorithm and incrementally improve upon it. So on with it. So the complexity of the algorithm is O(n^2). We can do this even faster using divide and conquer, more aptly if we use a variation of Merge Sort . To do this, unfortunately, we will have to mutate the array. The key to the algorithm is to understand that in an ascending sorted array the inversion count is zero. Say [10,20,30] If we perhaps have an element 100 before the rest of the array, the inversion count becomes 3. [100,10,20,30]. If we placed the 100 at the second position the inversi