Implement Queue using Stacks

Clarify the problem:

• The problem requires implementing a queue using stacks.
• We need to implement the functionality of a queue, including the push, pop, peek, and empty operations, using only stacks.

Analyze the problem:

• Input: None or the elements to be pushed into the queue.
• Output: The result of the operations (e.g., the popped element, the front element, or a boolean indicating whether the queue is empty).
• Constraints:
  • All operations should be implemented using stacks.
  • The input elements may be of any data type.
  • The queue can handle any number of elements.

Design an algorithm:

• We can implement a queue using two stacks.
• We will use one stack, called stack1, to store the elements in the order they are pushed.
• When we need to perform a pop or peek operation, we will transfer the elements from stack1 to another stack, called stack2, reversing their order.
• By doing this, the top element of stack2 will represent the front of the queue, allowing us to perform the required operations.
• If we push new elements while stack2 is not empty, we will transfer the elements back to stack1 before pushing the new elements.
• This ensures that the order of the elements is maintained.

Explain your approach:

• We will implement a class called MyQueue to represent the queue using stacks.
• The class will have two stacks, stack1 and stack2, as instance variables.
• The push operation will simply push the element into stack1.
• The pop operation will transfer the elements from stack1 to stack2 if stack2 is empty, and then pop the top element from stack2.
• The peek operation will be similar to pop, but it will only return the top element of stack2 without removing it.
• The empty operation will return True if both stack1 and stack2 are empty, indicating that the queue is empty.

Write clean and readable code:

python

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

    def push(self, x):
        self.stack1.append(x)

    def pop(self):
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2.pop()

    def peek(self):
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2[-1]

    def empty(self):
        return len(self.stack1) == 0 and len(self.stack2) == 0

Test your code:

• To test the code, we can create an instance of MyQueue and perform various operations on it.

python

# Test case:
q = MyQueue()
q.push(1)
q.push(2)
q.push(3)
print(q.pop())  # Output: 1
print(q.peek())  # Output: 2
print(q.empty())  # Output: False

Optimize if necessary:

• The implementation provided is already efficient, as it achieves the required operations in constant time complexity.
• No further optimizations are necessary.

Handle error cases:

• The code does not handle any explicit error cases.
• We assume that the user will only call the operations defined in the class and provide valid inputs.

Discuss complexity analysis:

• Time complexity:
  • The push operation has a time complexity of O(1) as it appends the element to stack1.
  • The pop and peek operations have an amortized time complexity of O(1).
    • In the worst case, when stack2 is empty, we need to transfer all elements from stack1 to stack2, which takes O(n) time, where n is the number of elements in the queue. However, this transfer occurs only once for every n elements, resulting in an amortized time complexity of O(1) per operation.
  • The empty operation has a time complexity of O(1) as it checks the lengths of stack1 and stack2.
• Space complexity:
  • The space complexity is O(n), where n is the number of elements in the queue.
  • Both stack1 and stack2 can potentially store all n elements, resulting in a linear space complexity.
  • However, since we transfer elements between the stacks as needed, the actual space used is determined by the number of elements being operated on, resulting in efficient space utilization.