Summary
This article is for beginners. It introduces the following ideas: Stack, Queue.
Solution
Approach #1 (Two Queues, push  , pop )
Intuition
Stack is LIFO (last in  first out) data structure, in which elements are added and removed from the same end, called top
.
In general stack is implemented using array or linked list, but in the current article we will review a different approach for implementing stack using queues. In contrast queue is FIFO (first in  first out) data structure, in which elements are added only from the one side  rear
and removed from the other  front
. In order to implement stack using queues, we need to maintain two queues q1
and q2
. Also we will keep top stack element in a constant memory.
Algorithm
Push
The new element is always added to the rear of queue q1
and it is kept as top
stack element
Figure 1. Push an element in stack
Java
private Queue<Integer> q1 = new LinkedList<>(); private Queue<Integer> q2 = new LinkedList<>(); private int top; // Push element x onto stack. public void push(int x) { q1.add(x); top = x; }
Complexity Analysis

Time complexity : . Queue is implemented as linked list and
add
operation has time complexity. 
Space complexity :
Pop
We need to remove the element from the top of the stack. This is the last inserted element in q1
.
Because queue is FIFO (first in  first out) data structure, the last inserted element could be removed only after all elements, except it, have been removed. For this reason we need to maintain additional queue q2
, which will serve as a temporary storage to enqueue the removed elements from q1. The last inserted element in q2
is kept as top. Then the algorithm removes the last element in q1
. We swap q1
with q2
to avoid copying all elements from q2
to q1
.
Figure 2. Pop an element from stack
Java
// Removes the element on top of the stack. public void pop() { while (q1.size() > 1) { top = q1.remove(); q2.add(top); } q1.remove(); Queue<Integer> temp = q1; q1 = q2; q2 = temp; }
Complexity Analysis
 Time complexity : . The algorithm dequeues n elements from
q1
and enqueues elements toq2
, where is the stack size. This gives operations.  Space complexity : .
Approach #2 (Two Queues, push  , pop )
Algorithm
Push
The algorithm inserts each new element to queue q2
and keep it as the top
element. In case queue q1
is not empty (there are elements in the stack), we remove all elements from q1
and add them to q2
. In this way the new inserted element (top
element in the stack) will be always positioned at the front of q2
. We swap q1
with q2
to avoid copying all elements from q2
to q1
.
Figure 3. Push an element in stack
Java
public void push(int x) { q2.add(x); top = x; while (!q1.isEmpty()) { q2.add(q1.remove()); } Queue<Integer> temp = q1; q1 = q2; q2 = temp; }
Complexity Analysis

Time complexity : . The algorithm removes n elements from
q1
and inserts elements toq2
, where n is the stack size. This gives operations. The operationsadd
andremove
in linked lists has complexity. 
Space complexity : .
Pop
The algorithm dequeues an element from queue q1
and keeps front element of q1
as top
.
Figure 4. Pop an element from stack
Java
// Removes the element on top of the stack. public void pop() { q1.remove(); if (!q1.isEmpty()) { top = q1.peek(); } }
Complexity Analysis
 Time complexity : .
 Space complexity : .
In both approaches empty
and top
operations have the same implementation.
Empty
Queue q1
always contains all stack elements, so the algorithm checks q1
size to return if the stack is empty.
// Return whether the stack is empty. public boolean empty() { return q1.isEmpty(); }
Time complexity : .
Space complexity : .
Top
The top
element is kept in constant memory and is modified each time when we push or pop an element.
// Get the top element. public int top() { return top; }
Time complexity : .
The top
element has been calculated in advance and only returned in top
operation.
Space complexity : .
Approach #3 (One Queue, push  , pop )
The mentioned above two approaches have one weakness, they use two queues. This could be optimized as we use only one queue, instead of two.
Algorithm
Push
When we push an element into a queue, it will be stored at back of the queue due to queue's properties. But we need to implement a stack, where last inserted element should be in the front of the queue, not at the back. To achieve this we can invert the order of queue elements when pushing a new element.
Figure 5. Push an element in stack
Java
private LinkedList<Integer> q1 = new LinkedList<>(); // Push element x onto stack. public void push(int x) { q1.add(x); int sz = q1.size(); while (sz > 1) { q1.add(q1.remove()); sz; } }
Complexity Analysis

Time complexity : . The algorithm removes n elements and inserts elements to
q1
, where n is the stack size. This gives operations. The operationsadd
andremove
in linked lists has complexity. 
Space complexity : .
Pop
The last inserted element is always stored at the front of q1
and we can pop it for constant time.
Java
// Removes the element on top of the stack. public void pop() { q1.remove(); }
Complexity Analysis
 Time complexity : .
 Space complexity : .
Empty
Queue q1
contains all stack elements, so the algorithm checks if q1
is empty.
// Return whether the stack is empty. public boolean empty() { return q1.isEmpty(); }
Time complexity : .
Space complexity : .
Top
The top
element is always positioned at the front of q1
. Algorithm return it.
// Get the top element. public int top() { return q1.peek(); }
Time complexity : .
Space complexity : .
Analysis written by: @elmirap.