232 Implement Queue using Stack - Easy

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.

Problem Analysis:

1. Time Complexity:
   • push(x): O(1) - Appending an element to the end of a list in Python has constant time complexity on average.
   • pop(): O(n) - Removing the first element from a list requires shifting all remaining elements to fill the gap. This results in linear time complexity, where "n" is the number of elements in the list.
   • peek(): O(1) - Accessing the first element of the list is a constant time operation.
   • empty(): O(1) - Checking the length of the list and comparing it with 0 is a constant time operation.
2. Space Complexity:
   • The space complexity is O(n), where "n" is the number of elements in the queue. This is because the elements are stored in a list.

Solutions:

class MyQueue:

    def __init__(self):
        self.queue = []

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

    def pop(self) -> int:
        return self.queue.pop(0)
        

    def peek(self) -> int:
        return self.queue[0]
        

    def empty(self) -> bool:
        return len(self.queue) == 0


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

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