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What is the minimum number of stacks of size $n$ required to implement a queue of size $n$?

1. One
2. Two
3. Three
4. Four

edited by

but we can also implement queue using single stack if other constraints also givenhttp://gateoverflow.in/2007/gate2014-2-41

set2018

It may look like only one stack is required but REVERSE operation can't be done without another stack.

Therefore minimum 2 stacks are required.

yes for this question it will 2 .if reverse is already given then 1 stack is sufficient

what do you mean by if the reverse is already given??

In this question, queue implemented using the stack.

maintaining the queue property, First-In-First-Out (or) Last-In-Last-Out, we need a minimum of two stacks.

bottom of one stack = top of another stack

@Lakshman Patel RJIT

>can't we use recursion for reversing?then only one stack excluding System stack isn't?

>min Queue to imp Stack will also be same right?

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A queue can be implemented using two stacks.

Let queue be represented as " $q$ "
and stacks used to implement $q$ be "stack1" and "stack2".

$q$ can be implemented in two ways:

Method 1 (By making EnQueue operation costly)

This method makes sure that newly entered element is always at the bottom of stack 1, so that deQueue operation just pops from stack1. To put the element at top of stack1, stack2 is used.

$enQueue(q, x)$

1. While stack1 is not empty, push everything from stack1 to stack2.
2. Push $x$ to stack1 (assuming size of stacks is unlimited).
3. Push everything back to stack1.

$dnQueue(q)$

1. If stack1 is empty then error
2. Pop an item from stack1 and return it

Method 2 (By making deQueue operation costly)

In this method, in en-queue operation, the new element is entered at the top of stack1. In de-queue operation, if stack2 is empty then all the elements are moved to stack2 and finally top of stack2 is returned.
$enQueue(q, x)$

1. Push $x$ to stack1 (assuming size of stacks is unlimited).

$deQueue(q)$

1. If both stacks are empty then error.
2. If stack2 is empty
While stack1 is not empty, push everything from stack1 to stack2.
3. Pop the element from stack2 and return it.

Correct Answer: $B$

by

in method 1), newly entered element is being entered at bottom of stack1, not on top. on top there is most old element which being popped first.
Corrected. Thanks :)
ans b)

### 1 comment

is it a theory or any explanation is present?
Two stacks we need to implement one queue

Two stacks of size n required to implement a queue of size n

Just one stack is needed to implement queue.

If stack is implemented using single link list then, we can traverse till the bottom of stack and return the value which should be returned for pop() operation.