# Implement Stack Using Queues

Companies:

Companies: Microsoft

Implement the following operations of a stack using queues.

push(x) — Push element x onto stack.
pop() — Removes the element on top of the stack.
top() — Get the top element.
empty() — Return whether the stack is empty.

#### Example:

MyStack stack = new MyStack();

stack.push(6);

stack.push(3);

stack.top(); // returns 3

stack.pop(); // returns 3

stack.top(); // returns 6

stack.empty(); // returns false

Notes:

• You must use only standard operations of a queue — which means only push to back, peek/pop from front, size, and is empty operations are valid.
• Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

### Solution:

Stack is based on LIFO i.e. Last In First Out, adding and removing of an element is done from the same end which is the top of the stack. We add the element at the top of the stack and pop it from the same end only. While, Queue is based on FIFO i.e. First In First Out, adding of the element is done from the rear side and removing of an element is done from the front.

A simple implementation is done using two queues. But we will go for more optimal solution and taking just one queue we can solve this problem more efficiently.

When we add an element to the queue, it get’s added from the rear side and when we remove an element, the element at front is poped out first. This is totally opposite to the case of stack.

To convert a queue into a stack without using any extra queue, we need to rotate the elements everytime we perform push operation, such that before any other next operation gets performed, it gets converted into a stack.

### Java:

``````class Solution {
class MyStack {

// Push element x onto stack.
public void push(int x) {
q.offer(x);
for (int i = 1; i < q.size(); i++) {
q.offer(q.remove());
}
}

// Removes the element on top of the stack.
public void pop() {
q.poll();
}

// Get the top element.
public int top() {
return q.peek();
}

// Return whether the stack is empty.
public boolean empty() {
return q.isEmpty();
}
}
}
``````

### Complexity Analysis:

• Time Complexity: O(n), for push() function.
• Space Complexity: O(1).
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