Given a stack of integers. The task is to design a special stack such that maximum element can be found in O(1) time and O(1) extra space.
Given Stack : 2 5 1 64 --> Maximum So Output must be 64 when getMax() is called.
Below are the different functions designed to push and pop elements from the stack.
Push(x) : Inserts x at the top of stack.
- If stack is empty, insert x into the stack and make maxEle equal to x.
- If stack is not empty, compare x with maxEle. Two cases arise:
- If x is less than or equal to maxEle, simply insert x.
- If x is greater than maxEle, insert (2*x – maxEle) into the stack and make maxEle equal to x. For example, let previous maxEle was 3. Now we want to insert 4. We update maxEle as 4 and insert 2*4 – 3 = 5 into the stack.
Pop() : Removes an element from top of stack.
- Remove element from top. Let the removed element be y. Two cases arise:
- If y is less than or equal to maxEle, the maximum element in the stack is still maxEle.
- If y is greater than maxEle, the maximum element now becomes (2*maxEle – y), so update (maxEle = 2*maxEle – y). This is where we retrieve previous maximum from current maximum and its value in stack. For example, let the element to be removed be 5 and maxEle be 4. We remove 5 and update maxEle as 2*4 – 5 = 3.
- Stack doesn’t hold actual value of an element if it is maximum so far.
- Actual maximum element is always stored in maxEle
- Number to be Inserted: 3, Stack is empty, so insert 3 into stack and maxEle = 3.
- Number to be Inserted: 5, Stack is not empty, 5> maxEle, insert (2*5-3) into stack and maxEle = 5.
- Number to be Inserted: 2, Stack is not empty, 2< maxEle, insert 2 into stack and maxEle = 5.
- Number to be Inserted: 1, Stack is not empty, 1< maxEle, insert 1 into stack and maxEle = 5.
- Initially the maximum element maxEle in the stack is 5.
- Number removed: 1, Since 1 is less than maxEle just pop 1. maxEle=5.
- Number removed: 2, 2<maxEle, so number removed is 2 and maxEle is still equal to 5.
- Number removed: 7, 7> maxEle, original number is maxEle which is 5 and new maxEle = 2*5 – 7 = 3.
Number Inserted: 3 Number Inserted: 5 Maximum Element in the stack is: 5 Number Inserted: 7 Number Inserted: 19 Maximum Element in the stack is: 19 Top Most Element Removed: 19 Maximum Element in the stack is: 7 Top Most Element Removed: 7 Top Most Element is: 5
- Design a stack that supports getMin() in O(1) time and O(1) extra space
- Clone a stack without extra space
- Reverse a stack without using extra space in O(n)
- Design a stack to retrieve original elements and return the minimum element in O(1) time and O(1) space
- Find duplicate in an array in O(n) and by using O(1) extra space
- Find maximum in stack in O(1) without using additional stack
- Sum of K largest elements in BST using O(1) Extra space
- Design and Implement Special Stack Data Structure | Added Space Optimized Version
- Count Fibonacci numbers in given range in O(Log n) time and O(1) space
- Maximum removal from array when removal time >= waiting time
- Tracking current Maximum Element in a Stack
- Infix to Postfix using different Precedence Values for In-Stack and Out-Stack
- Stack Permutations (Check if an array is stack permutation of other)
- Stack | Set 3 (Reverse a string using stack)
- Sort a stack using a temporary stack
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : princiraj1992