232. Implement Queue using Stacks

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

  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.

Solution

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

/** 
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function(x) {
    this.stack1.push(x);
};

/**
 * @return {number}
 */
MyQueue.prototype.pop = function() {
    if (this.stack2.length === 0) {
        while (this.stack1.length) this.stack2.push(this.stack1.pop());
    }
    return this.stack2.pop();
};

/**
 * @return {number}
 */
MyQueue.prototype.peek = function() {
    if (this.stack2.length === 0) {
        while (this.stack1.length) this.stack2.push(this.stack1.pop());
    }
    return this.stack2[this.stack2.length - 1];
};

/**
 * @return {boolean}
 */
MyQueue.prototype.empty = function() {
    return this.stack1.length === 0 && this.stack2.length === 0;
};

/** 
 * Your MyQueue object will be instantiated and called as such:
 * var obj = new MyQueue()
 * obj.push(x)
 * var param_2 = obj.pop()
 * var param_3 = obj.peek()
 * var param_4 = obj.empty()
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).