Lost in Compilation

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)

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 ...

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 integers. Find the longest increasing subsequence in the array. Eg. [42, 91, 46, 73, 75, 91, 95]. The LIS for this array is 6 which is [42,46,73,75,91,95]. Note: Subsequence does not require the elements to be contiguous As always, we begin with our naive solution which is a DFS/Bruteforce search trying every possible combination until we get the longest sequence. This is simply exhaustive and will explode on bigger arrays. Time Complexity : O(2^n) ~ Exponential Now, If you notice we repeatedly try a prefix subsequence repeatedly. This tells us, the problem is a good candidate for a dynamic programming algorithm. We define lis[0...L] , where L is the length of the array. We define lis[i] as the length of longest increasing subsequence that ends at i.  lis[0] = 1 { trivial since, the longest increasing sequence that ends at the start of the array contains just the first element, also true for the arr[i] where arr[i] is the smaller th...