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

Shortest Interval in k-sorted list - Hard

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)
Post a Comment
Read more

The 2 Missing Duplicate Numbers - Medium

This is a slightly trickier version of The Missing Duplicate Number . Instead of one single missing number, we have 2. Problem : Given an array of Integers. Each number is repeated even number of times. Except 2 numbers which occur odd number of times. Find the two numbers Ex. [1,2,3,1,2,3,4,5] The Output should be 4,5. The naive solution is similar to the previous solution. Use a hash table to keep track of the frequencies and find the elements that occur an odd number of times. The algorithm has a Time and Space Complexity of O(n). Say those two required numbers are a and b If you remember the previous post, we used the XOR over all the elements to find the required element. If we do the same here, we get a value xor which is actually a ^ b. So how can we use this? We know that a and b are different, so their xor is not zero. Infact their XOR value has its bit set for all its dissimilar bits in both a and b. (eg.  1101 ^ 0110 = 0011). It's important to notice ...
Post a Comment
Read more