How can you implement a Queue using two stacks?

Implement the following operations of a queue using 2 stacks named s1 and s2.

  • push(x) → Push element x to the back of the Queue.
  • pop() → Removes the element from in front of the Queue.
  • peek() → Get the front element.
  • empty() → Return whether the Queue is empty.

Notes:

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

The enQueue algorithm is very simple, Just push the value on the stack s1. The time complexity of enQueue operation is O(1).

deQueue Algorithm

  • If the stack s2 is not empty then pop from s2 and return the element.
  • If the stack s2 is empty, then transfer all elements from s1 to s2. And return the popped element from s2. We can optimize the code a little by transferring only (n-1) elements from stack s1 to stack s2 and pop the nth element from stack s2 return and the poped element. 
  • If stack s1 is also empty then throw an error.

The time complexity of this algorithm, if the stack s2 is not empty then the complexity is O(1). But if the stack s2 is empty, then we need to transfer all the elements from s1 to s2. But if we carefully observe, the number of transferred elements and the number of popped elements from s2 are equal. Due to this average complexity of pop operation, in this case, is O(1). The amortized complexity of pop operation is O(1). The following figure is the example of this algorithm.

class MyQueue {

  private Stack<Integer> enQueueStack;
  private Stack<Integer> deQueueStack;

  /** Initialize your data structure here. */
  public MyQueue() {
    enQueueStack = new Stack<>();
    deQueueStack = new Stack<>();
  }


  /** Push element x to the back of queue. */
  public void push(int x) {
    enQueueStack.push(x);
  }

  /** Removes the element from in front of queue and returns that element. */
  public int pop() {
    Integer item = null;
    if (!deQueueStack.isEmpty()) {
      item = deQueueStack.pop();
    } else {
      while (!enQueueStack.isEmpty()) {
        deQueueStack.push(enQueueStack.pop());
      }
      item = deQueueStack.pop();
    }
    return item;
  }

  /** Get the front element. */
  public int peek() {
    if (!deQueueStack.isEmpty()) {
      return deQueueStack.peek();
    } else {
      while (!enQueueStack.isEmpty()) {
        deQueueStack.push(enQueueStack.pop());
      }
      return deQueueStack.peek();
    }
  }

  /** Returns whether the queue is empty. */
  public boolean empty() {
    return enQueueStack.isEmpty() && deQueueStack.isEmpty();
  }
}

Amortized Analysis:

The amortized analysis gives the average performance (over time) of each operation in the worst case. The basic idea is that a worst-case operation can alter the state in such a way that the worst case cannot occur again for a long time, thus amortizing its cost. Consider this example where we start with an empty queue with the following sequence of operations applied:

push1,push2,…,pushn,pop1,pop2…,popn 

The worst-case time complexity of a single pop operation is O(n). Since we have n pop operations, using the worst-case per operation analysis gives us a total of O(n2) time. However, in a sequence of operations the worst case does not occur often in each operation - some operations may be cheap, some may be expensive. Therefore, a traditional worst-case per operation analysis can give overly pessimistic bound. For example, in a dynamic array, only some inserts take linear time, though others - a constant time. In the example above, the number of times pop operation can be called is limited by the number of push operations before it. Although a single pop operation could be expensive, it is expensive only once per n times (Queue sSze), when s2 is empty and there is a need for data transfer between s1 and s2. Hence the total time complexity of the sequence is : 

n (for push operations) + 2*n (for first pop operation) + n - 1 ( for pop operations) which is O(2*n).

This gives O(2n/2n) = O(1) average time per operation.

Algorithm