Implement Queue using Stacks

Implement Queue using Stacks - LeetCodeLeetCode

Problem

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.
 

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output : [null, null, null, 1, 1, false]

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

Pseudocode

a stack is last in first out (LIFO) while a queue is first in first out (FIFO)
- behavior of a stack,
    - can only use push() and pop(), append last element
- implement a function that obtains the first element pushed i.e. shift() in JS, using two stacks while only using pop() and push()
- two stacks are provided
    - divide the two stacks into 
        - input stack, where all the values are pushed into
        - output stack, where the top most stack is the first value pushed
            - when the first is requested
            - pop() input stack and push() into output stack, until output in empty
            - the topmost value in output stack is the first value pushed in
            - to maintain ordering, pop everything out of output stack into intput

Solution

var MyQueue = function () {
  this.stack1 = [];
  this.stack2 = [];
};

/**
 * Push element x to the back of queue.
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function (x) {
  this.stack1.push(x);
};

/**
 * Removes the element from in front of queue and returns that element.
 * @return {number}
 */
MyQueue.prototype.pop = function () {
  while (this.stack1.length !== 0) {
    this.stack2.push(this.stack1.pop());
  }

  var pop = this.stack2.pop();

  while (this.stack2.length !== 0) {
    this.stack1.push(this.stack2.pop());
  }

  return pop;
};

/**
 * Get the front element.
 * @return {number}
 */
MyQueue.prototype.peek = function () {
  while (this.stack1.length !== 0) {
    this.stack2.push(this.stack1.pop());
  }

  var pop = this.stack2.pop();
  this.stack2.push(pop);
  while (this.stack2.length !== 0) {
    this.stack1.push(this.stack2.pop());
  }

  return pop;
};

/**
 * Returns whether the queue is empty.
 * @return {boolean}
 */
MyQueue.prototype.empty = function () {
  return this.stack1.length === 0 ? true : false;
};

Time and Space Complexity

Time

• pushing and popping elements out on an array is constant time operation O(1)

• retrieval of element from FIFO requires the stack to be pop and pushed for the entire array length O(N)

• Total - O(N)

Space

• Memory required to store elements pushed in is O(N)

• Total - O(N)