What are the operations perform on stack?

What are the operations perform on stack?

Mainly the following three basic operations are performed in the stack:

• Push: Adds an item in the stack. If the stack is full, then it is said to be an Overflow condition.
• Pop: Removes an item from the stack.
• Peek or Top: Returns top element of stack.
• isEmpty: Returns true if stack is empty, else false.

How do you do push operations in stack?

Push Operation

1. Step 1 − Checks if the stack is full.
2. Step 2 − If the stack is full, produces an error and exit.
3. Step 3 − If the stack is not full, increments top to point next empty space.
4. Step 4 − Adds data element to the stack location, where top is pointing.
5. Step 5 − Returns success.

What are the operations of queue?

Mainly the following four basic operations are performed on queue: Enqueue: Adds an item to the queue. If the queue is full, then it is said to be an Overflow condition. Dequeue: Removes an item from the queue.

How a stack can be implemented using two queues?

Likewise, a queue can be implemented with two stacks, a stack can also be implemented using two queues. The basic idea is to perform stack ADT operations using the two queues. So, we need to implement push(),pop() using DeQueue(), EnQueue() operations available for the queues.

What should the stack look like when getminimum is called?

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. So after this, the stack looks like

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.

How to return the minimum element of a stack?

Design a Stack such that the operation of getminimum () (function returning minimum element of the stack) also takes constant time. Please note that it will only return the current minimum element from the stack and will not delete (pop) any element from the stack. minStack to hold the current minimum element.

How to solve the problem of Min stack?

We can solve this problem of min stack by using two stacks. i) Declare two stacks. One is the main stack in which we push value as it is. In the second stack, we only push the minimum element present at that time. ii) Whenever we perform push operation in a stack. a) Push the value as it is in a first stack.