Lost in Compilation

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]. So, 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).

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 simila...

Problem - Given an Array of integers, A. Find out if there exists a triple (i,j,k) such that A[i] + A[j] + A[k] == 0. The 3SUM  problem is very similar to the 2SUM  problem in many aspects. The solutions I'll be discussing are also very similar. I highly recommend you read the previous post first, since I'll explain only the differences in the algorithm from the previous post. Let's begin, We start with the naive algorithm. An O(n^3) solution with 3 nested loops each checking if the sum of the triple is 0. Since O(n^3) is the higher order term, we can sort the array in O(nlogn) and add a guard at the nested loops to prune of parts of the arrays. But the complexity still remains O(n^3). The code is pretty simple and similar to the naive algorithm of 2SUM. Moving on, we'll do the same thing we did in 2SUM, replace the inner-most linear search with a binary search. The Complexity now drops to O(n^2 logn) Now, the hash table method, this is strictly not ...

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 ...

As promised, this is the follow up to the first posted earlier that there was a more efficient algorithm for finding the longest increasing subsequence, better than the O(n^2) DP solution we came up with. It's not easy to understand, but once you get what the variables mean, it seems pretty straight forward. We'll be implementing another DP solution, similar to the previous one, except with an additional data structure that'll reduce the overall complexity to O(nlogn). The wikipedia seems no help. The explanation is contrived and not easy to visualize. Hope this post will clear a lot of doubts. So we need 3 arrays for solving an instance of LIS. M[i] - where M[i] is the minimum ending element of a increasing subsequence of length i. This is the key part of the new algorithm. Unlike the earlier solution where we kept the longest length discovered at index i , we store an Array which holds the minimum value of a subsequence of length i. I[i] - where M[i] = A[I[i]] si...

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...