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QuickSelect - Medium

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 algorithm us…
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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, [10,10…

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 origi…

Find Increasing Triplet Subsequence - Medium

Problem - Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j < k <= n and A[i] < A[j] < A[k].
Let's start with the obvious solution, bruteforce every three element combination until we find a solution.

Let's try to reduce this by searching left and right for each element, we search left for an element smaller than the current one, towards the right an element greater than the current one. When we find an element that has both such elements, we found the solution. The complexity of this solution is O(n^2).
To reduce this even further, One can simply apply the longest increasing subsequence algorithm and break when we get an LIS of length 3. But the best algorithm that can find an LIS is O(nlogn) with O( n ) space.

An O(nlogn) algorithm seems like an overkill! Can this be done in linear time?

The Algorithm:
We iterate over the array and keep track of two things.
The minimum value iterated over (min)The minimum increasing pair (a,b) where…

Find the Quadruplets - Hard

Problem - Given 4 arrays A,B,C,D. Find out if there exists an instance where A[i] + B[j] + C[k] + D[l] = 0
Like the Find the Triple problem, we're going to develop 4 algorithms to solve this. Starting with the naive O(n^4) solution.
Then we proceed to eliminate the inner-most loop with a Binary Search, reducing the complexity to O(n^3 logn)

Now, we replace the last 2 loops with the left-right traversal we did in the previous 3 posts.

Now the complexity is O(n^3).

Finally, we reduce the complexity to O(n^2 logn) at the cost of O(n^2) Space Complexity. We store every combination of A[i] + B[j] and store it in AB[]. Similarly we make CD[] out of C[i] + D[j].


AB = A x B
CD = C x D

We then sort AB and CD (which costs O(n^2 log(n^2)) ~ O(n^2 logn) ) and then run a left-right linear Algorithm on AB and CD. (Note : Their size is of the order O(n^2))

So the overall complexity is due to sorting the large array of size n^2. which is O(n^2 logn).

Find the Triple - Hard

Problem : Given 3 Integer Arrays A,B,C. Find out if an instance exists where A[i] + B[j] + C[k] == 0.
Very deceptive question but very similar to the 3SUM Problem. The Naive solution is O(n^3). If you're a little clever, you can do a O(n^2 logn). At this point, you might give up assuming that this is pretty efficient. But then there is also a O(n^2) solution as well!

Let's go from worst to best.

The Naive solution is trivial.
Try every combination until we hit a solution, if we don't there is no solution.

So let's try to speed things up a bit, since the solution is already O(n^3), I think we can afford to sort any of the  arrays since nlogn = o(n^3). Therefore, we cleverly sort the Array C. Now in the third loop we replace the linear scan with a binary search to find (-A[i]-B[j]). The complexity is now reduced to O(n^2 logn)

At this point, you've leveled up. The next algorithm might not be immediately apparent but works. In fact the method is similar to the O(n^2)…