Skip to main content

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 min <= a < b
When we encounter an element, say v,  that is greater than the increasing pair we return (a,b,v).

Here's my Python implementation to make my point.

Until next time, Cheers!
Labels:

Comments

  1. UnknownOctober 30, 2013 at 9:46 PM

    Good job, I really like the way you explained. For the last implementation, I think you should change to "elif b == None or b >= v" otherwise it will fail on [3,2,4,4]

    ReplyDelete
  2. st0leNovember 1, 2013 at 2:52 PM

    It has to fail for [3,2,4,4]...The question clearly states the triple should be strictly increasing.

    ReplyDelete

Post a Comment

Popular posts from this blog

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 ...
Post a Comment
Read more

3SUM - Hard

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 ...
Post a Comment
Read more