Skip to main content

Stack with Min() Operation - Easy

Problem : Implement a stack with push() , pop() and a min() (which returns the minimum element in the stack). All operations should be O(1).
Another popular question, I've searched around a lot came across many answers, but only one had completeness. It's important to note that with the stack you don't have random access to the contents of the stack when implementing your solution.

The solution I'm describing involves using 2 stacks, one is to store the data elements themselves (mainstack) and the other to store minimum values, which i'll refer to as minstack. Now we can have 2 parallel stacks where we push and pop on the minstack along with the mainstack, keeping track of the minimum value of course. But we can optimize this further. We only need to push values on the minimum stack only when you push a value on the main stack that is less than or equal to the current top element of the minstack.
Here's are the two stacks after pushing 67,44,20,77,99,12 in that order.



When an element is poped, you check if the value on the minstack top is the same as the one on the mainstack, if it is we pop it from the minstack as well. Here's me poping 12.


Let's get to the code now. I pick Java as my weapon of choice.

Complete Code can be found here. It's important to note that we can implement both max() and min() for the same stack. You'll need one more stack for keeping track of the maximum element. I'll leave that as an exercise for you. Cheers!
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

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

Inversion Count - Medium

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 inversi
Post a Comment
Read more