Implement Queue using Stacks Python Solution

LeetCode 232, Easy. Topics: Stack, Design, Queue. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 12 parametrized pytest cases, and a playground notebook:

lcpy gen -n 232   # by problem number
lcpy gen -s implement_queue_using_stacks   # by problem name

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.

Examples

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.

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.

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

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class MyQueue:
    # Time: O(1)
    # Space: O(n)
    def __init__(self) -> None:
        self.input_stack: list[int] = []
        self.output_stack: list[int] = []

    # Time: O(1)
    # Space: O(1)
    def push(self, x: int) -> None:
        self.input_stack.append(x)

    # Time: O(1) amortized
    # Space: O(1)
    def pop(self) -> int:
        self._move_to_output()
        return self.output_stack.pop()

    # Time: O(1) amortized
    # Space: O(1)
    def peek(self) -> int:
        self._move_to_output()
        return self.output_stack[-1]

    # Time: O(1)
    # Space: O(1)
    def empty(self) -> bool:
        return not self.input_stack and not self.output_stack

    def _move_to_output(self) -> None:
        if not self.output_stack:
            while self.input_stack:
                self.output_stack.append(self.input_stack.pop())


# Amortized O(1) Explanation:
# Example with 4 push + 4 pop operations:
#
# push(1)  # input: [1], output: []           - O(1)
# push(2)  # input: [1,2], output: []         - O(1)
# push(3)  # input: [1,2,3], output: []       - O(1)
# push(4)  # input: [1,2,3,4], output: []     - O(1)
#
# pop()    # Move all 4 to output: input: [], output: [4,3,2,1] then pop 1  - O(4)
# pop()    # output: [4,3,2], just pop 2                                     - O(1)
# pop()    # output: [4,3], just pop 3                                       - O(1)
# pop()    # output: [4], just pop 4                                         - O(1)
#
# Total cost: 4 + 4 + 1 + 1 + 1 = 11 operations for 8 calls = 1.4 per operation
# Key: Each element moves exactly once from input to output, so expensive O(n)
# transfer is "spread out" over multiple cheap O(1) operations = amortized O(1)

Complexity

Time	Space
O(1)	O(n)

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Implement Queue using Stacks Python Solution

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