# 232. Implement Queue using Stacks

## 1. Question

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

## 2. Examples

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], , , [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: 
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is 
myQueue.empty(); // return false


## 3. Constraints

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid.

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

## 5. Solutions

class MyQueue {

Stack<Integer> sa;
Stack<Integer> sb;

public MyQueue() {
sa = new Stack<>();
sb = new Stack<>();
}

public void push(int x) {
while (!sb.isEmpty()) {
sa.push(sb.pop());
}
sa.push(x);
}

public int pop() {
return getFirst(true);
}

public int peek() {
return getFirst(false);
}

public boolean empty() {
while (!sb.isEmpty()) {
sa.push(sb.pop());
}
return sa.isEmpty();
}

private int getFirst(boolean needPop) {
while (!sa.isEmpty()) {
sb.push(sa.pop());
}
int a = needPop ? sb.pop() : sb.peek();
return a;
}
}