Skip to main content

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.

Comments

Popular posts from this blog

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 increa...

The 2 Missing Duplicate Numbers - Medium

This is a slightly trickier version of The Missing Duplicate Number . Instead of one single missing number, we have 2. Problem : Given an array of Integers. Each number is repeated even number of times. Except 2 numbers which occur odd number of times. Find the two numbers Ex. [1,2,3,1,2,3,4,5] The Output should be 4,5. The naive solution is similar to the previous solution. Use a hash table to keep track of the frequencies and find the elements that occur an odd number of times. The algorithm has a Time and Space Complexity of O(n). Say those two required numbers are a and b If you remember the previous post, we used the XOR over all the elements to find the required element. If we do the same here, we get a value xor which is actually a ^ b. So how can we use this? We know that a and b are different, so their xor is not zero. Infact their XOR value has its bit set for all its dissimilar bits in both a and b. (eg.  1101 ^ 0110 = 0011). It's important to notice ...

Matrix Rotation - Medium

Problem - Write an algorithm capable of rotating a matrix, in-place without any auxiliary storage. I don't have an elaborate thought process to develop this algorithm. It's a few rules you ought to remember to implement rotations in the matrix. To implement these rules, you need some helper functions which are quite simple really. You need to be able to transpose a matrix, reverse a row, reverse a column. That's it. The rest is just a combination of these 3 functions. Python is awesome, isn't it? Note : These functions create copies of the matrix, we can design algorithms that modify the original matrix with ease for square matrices. For non-square matrices, we have to create new matrices. Okay, let's get to the rotations. So we need to perform three kinds of rotations. 90,180,270. 1) Rotation by 90/-270  degrees Transpose the Matrix Reverse each row 2) Rotation by 180/-180 degrees There are two methods: First Method, is clearly obviou...