Skip to main content

Maximizing the Difference - Medium

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).
Labels:

Comments

Post a Comment

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...
2 comments
Read more

QuickSelect - Medium

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...
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