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Friday, August 29, 2014

Inversion Count - Medium

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 inversion count would be 2, [10,100,20,30].

So here's what we're going to do, we write a recursive function which returns inversion count. The function first recursively does it for the first half and second half of the array, the byproduct of these recursive calls is that both half's become sorted. (Just like Merge Sort!) Now if you have two sorted Arrays which should be merged into one, we can piggy back on the merging process (of merge sort!) and count the inversions by merely counting the number of times an element from the first half was greater than the element in the second half.

An example, for those who found my explanation confusing.

Say, at some point of the recursion, we have the following
First half : [20,40,60,80] , Second half : [10,30,50,70]

If you recall the merge subroutine, we first compare 20 and 10, since 20 > 10, that counts as an inversion. But also note, since the first half is sorted therefore it is guaranteed that  [20, 40, 60, 80] will all be greater than 10. Since 20 had to be the minimum element of the first half. Therefore the inversion count for 10 in this instance becomes 4 (size of the unexplored first half). Important to note, this algorithm only works if the integers are all distinct.

Speaking generally, let's say i is the index for first half, j is the index for the second. Aux is the auxillary array to store the result. Then,

If A[i] > A[j] then    // implies A[j] < A[i…mid]
    invCount = mid - i + 1
    Aux[k++] = A[j++]
Else
    Aux[k++] = A[i++]
End If

Now for the code, notice it's uncanny resemblance to merge sort.

And that's Inversion Count! :)

Friday, September 20, 2013

Shortest Interval in k-sorted list - Hard

Given k-sorted array, find the minimum interval such that there is at least one element from each array within the interval. Eg. [1,10,14],[2,5,10],[3,40,50]. Output : 1-3

To solve this problem, we perform a k-way merge as described here. At each point of 'popping' an element from an array. We keep track of the minimum and maximum head element (the first element) from all the k-lists. The minimum and maximum will obviously contain the rest of the header elements of the k-arrays. As we keep doing this, we find the smallest interval (max - min). That will be our solution. Here's a pictorial working of the algorithm.


And here's the Python code. Time Complexity : O(nlogk)


Merge k-sorted lists - Medium


Problem - Given k-sorted lists, merge them into a single sorted list.
A daft way of doing this would be to copy all the list into a new array and sorting the new array. ie O(n log(n))

The naive method would be to simply perform k-way merge similar to the auxiliary method in Merge Sort. But that is reduces the problem to a minimum selection from a list of k-elements. The Complexity of this algorithm is an abysmal O(nk).

Here's how it looks in Python. We maintain an additional array called Index[1..k] to maintain the head of each list.

We improve upon this by optimizing the minimum selection process by using a familiar data structure, the Heap! Using a MinHeap, we extract the minimum element from a list and then push the next element from the same list into the heap, until all the list get exhausted.

This reduces the Time complexity to O(nlogk) since for each element we perform O(logk) operations on the heap. An important implementation detail is we need to keep track of the origins of the elements of the heap, ie the list it came from and it's index. Therefore, I use a 3 member tuple (item, list-index, item-index) and push it onto the heap.

Fortunately for us, Python has the heapq module for implementing a heap. So here's the code.


The entire code can be found here.