How do you implement a stack which will support following operations in O 1 time complexity?
How to implement a stack which will support following operations in O(1) time complexity? 1) push() which adds an element to the top of stack. 2) pop() which removes an element from top of stack. 3) findMiddle() which will return middle element of the stack.
What is stack overflow condition *?
A stack overflow is an undesirable condition in which a particular computer program tries to use more memory space than the call stack has available. When a stack overflow occurs as a result of a program’s excessive demand for memory space, that program (and sometimes the entire computer) may crash.
What is Max stack?
The problem “Max stack” states to design a special stack which can perform these operations : push(x): push one element into the stack. top(): returns the element that is at the top of the stack. pop(): remove the element from the stack which is at the top. peekmax(): returns the maximum element of the stack.
What is the minimum number of queues needed to implement a priority queue?
What is minimum number of queues needed to implement priority queue? Two. One queue is used for actual storing of data and another for storing priorities.
How to design a stack to return the minimum element?
You need to design a stack which holds an integer value such that getMinimum () function should return the minimum element in the stack. When getMinimum () is called it should return 1, which is the minimum element in the stack. Note: Both 5 and 1 are popped out of the stack.
How to support minimum with O ( 1 ) time?
An approach that uses O (1) time and O (n) extra space is discussed here. In this article, a new approach is discussed that supports minimum with O (1) extra space. We define a variable minEle that stores the current minimum element in the stack.
When do you pop the main stack, pop the Min stack too?
When you pop the main stack, pop the min stack too. When you push the main stack, push either the new element or the current min, whichever is lower. getMinimum () is then implemented as just minStack.peek (). Please let me know if this isn’t enough information. It’s simple when you grok it, but it might take a bit of head-scratching at first 🙂
How to create a stack that supports getmin ( )?
Consider the following SpecialStack 16 –> TOP 15 29 19 18 When getMin () is called it should return 15, which is the minimum element in the current stack. If we do pop two times on stack, the stack becomes 29 –> TOP 19 18 When getMin () is called, it should return 18 which is the minimum in the current stack.