Lost in Compilation

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 in...

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 ...

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...

Problem : Given an array with distinct elements. Find out if there exists an instance where sum of two distinct elements is equal to a given constant K. Say, A = [1,2,3,4], K = 7. Output should be 3,4 The naive solution for this problem is pretty trivial. Brute force every pair-sum and check if the sum is K. You could be a little smart about this. You initially sort the array O(nlogn) and iterate the array pairwise and quit when the sum exceeds K, you can quit the inner loop since the array is sorted the sum can only increase. For the rest of the article, I'll be assuming K = 0, this doesn't affect the solutions in any way and it's trivial to introduce a new variable into the code without major changes. Here's the code, should be easy to follow... I'll be using Java today. Now that the naive solution is out of the way, let's try and do some clever stuff. So notice we can afford to sort the array since the brute solution was already O(n^2) (which ...

Problem: Given an array of positive non-zero integers. Find out the minimum hops required to traverse the array, if the value at an index denotes the maximum length you can hop from that index. To perform a brute force solution, we can use a depth first search with pruning. This usually explodes with arrays of size greater than 50. But never the less, it's important to understand how it works. Following our dismal performance using DFS, we move on to the big guns, Dynamic Programming. We define minHop[i]   as the minimum hops you need from index i to reach the end of the array. Therefore, minHop[i] = 1 { if arr[i] + i > len(arr) } needs just one hop, trivial. minHop[i] = 1 + min ( minHop[i+1 : i+arr[i]] ) { the minimum hop from my range which will reach the end of the array } We start with i = len(arr) - 1 and count down to i  = 0 . At which point minHop[0] will be the minimum hops required to reach the end of the array from index = 0. So here we hav...

Problem : In a party of N people, only one person is known to everyone. Such a person may be present in the party, if yes, (s)he doesn't know anyone in the party.  We can only ask questions like “does A know B? Find the celebrity in minimum number of questions. This is a very interesting problem. The answer, ie the least number of questions required, changes based on what's given. Are we guaranteed there's one celebrity in the party? Does every non-celeb know every other non-celeb? ( not very ideal, but under a best case scenario. ) Once again, the rules There is at most one celebrity at the party Everyone knows the celebrity. The celebrity knows no one. We can only ask the question "Does A know B?" When we ask the question, we can get a "yes" or "no". Each conveying information about both the participants in the question. Let's see those outcomes... If the Answer is "Yes", we know A is definitely not a celebrit...

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...

Problem : Given a set of integers A, find out if there is a subset of A which add up to a particular integer K. Formally called the Subset Sum Problem , we are given an NP-Complete problem. To solve this, we have to exhaustively search every subset and verify if they add up to an integer K. Let's begin with a Brute force DFS search, we're going to construct subsets incrementally until the sum exceeds K. The elements are assumed sorted in the increasing order. So we perform a DFS trying to perform every combination of the elements in the set. The recursive function either will include an element in the sum or exclude it. The guard prunes off DFS paths that exceed the sum required. There is an efficient method for this, using Dynamic Programming. Essentially, we start with a single element, add the next element together to get the new sums possible. These new sums will be the input for the next iteration. I'll explain with an example, say we have {a,b,c} You...

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 ...

Word ladder is a word game invented by Lewis Carroll. A word ladder puzzle begins with two words, and to solve the puzzle one must find a chain of other words to link the two, in which two adjacent words (that is, words in successive steps) differ by one letter. Eg. SPORT <-> SHORT Source: Wikipedia If you've followed my previous Dijkstra's Shortest Path Algorithm Tutorial, you'll be able to solve this challenge with minimum effort. Essentially, we create a graph where each word is a vertex and there exists an edge if two words differ by a single character. To find the word ladder is basically running any shortest path algorithm. You could also use a Breadth First Search. Some important observations The length of the source and destination words must be the same for a path to exist. The path from source to destination  can only contain words of the same length. So we can prune off the rest of the dictionary. The graph is undirected. Here's my ...

Problem : Given an array A of integers, maximize A[j] - A[i] such that 0 <= i < j < len(A). A very interesting problem, as always lets begin with a naive solution. Clearly, the solution has quadratic time complexity. Now, step back and look at the solution. Essentially what we are doing is iterating over the array, splitting the array into two parts. From the first part we always pick the lowest element, (clearly since that will give us the maximum difference). Since we already know the minimum element we encountered, we simply need to check if the difference with the new element improves upon the earlier solution. This idea can be implemented in Linear Time. Here's how. Update : So I just discovered this problem is called the Single Sell Profit problem. It's been discussed quite thoroughly by templatetypedef here . It includes 4 solutions (including this one).