232. Implement Queue using Stacks - Explanation

Problem Link

Description

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.


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1. Using Two Stacks (Brute Force)

class MyQueue:

    def __init__(self):
        self.stack1 = []
        self.stack2 = []

    def push(self, x: int) -> None:
        self.stack1.append(x)

    def pop(self) -> int:
        while len(self.stack1) > 1:
            self.stack2.append(self.stack1.pop())
        res = self.stack1.pop()
        while self.stack2:
            self.stack1.append(self.stack2.pop())
        return res

    def peek(self) -> int:
        while len(self.stack1) > 1:
            self.stack2.append(self.stack1.pop())
        res = self.stack1[-1]
        while self.stack2:
            self.stack1.append(self.stack2.pop())
        return res

    def empty(self) -> bool:
        return not self.stack1

Time & Space Complexity

  • Time complexity:
    • O(1)O(1) time for initialization.
    • O(1)O(1) time for each push()push() and empty()empty() function calls.
    • O(n)O(n) time for each pop()pop() and peek()peek() function calls.
  • Space complexity: O(n)O(n)

2. Using Two Stacks (Amortized Complexity)

class MyQueue:

    def __init__(self):
        self.s1 = []
        self.s2 = []

    def push(self, x: int) -> None:
        self.s1.append(x)

    def pop(self) -> int:
        if not self.s2:
            while self.s1:
                self.s2.append(self.s1.pop())
        return self.s2.pop()

    def peek(self) -> int:
        if not self.s2:
            while self.s1:
                self.s2.append(self.s1.pop())
        return self.s2[-1]

    def empty(self) -> bool:
        return max(len(self.s1), len(self.s2)) == 0

Time & Space Complexity

  • Time complexity:
    • O(1)O(1) time for initialization.
    • O(1)O(1) time for each push()push() and empty()empty() function calls.
    • O(1)O(1) amortized time for each pop()pop() and peek()peek() function calls.
  • Space complexity: O(n)O(n)