Queue

 Let's explain the concept of a queue, provide an example with proper comments, and solve a question from LeetCode that involves using a queue. We'll also analyze the time and space complexity of our solution.

  1. Queue: A queue is a data structure that follows the FIFO (First-In-First-Out) principle. It is similar to a real-life queue where the first person who joins the queue is the first one to be served.

In a queue, elements are added at the rear (enqueue) and removed from the front (dequeue). This ensures that the oldest elements are processed first.

  1. Example with Comments: Let's demonstrate the implementation of a queue with an example. We'll create a queue class in Python that supports the enqueue and dequeue operations.

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

    def enqueue(self, item):
        # Add the item to the rear of the queue
        self.queue.append(item)

    def dequeue(self):
        if not self.is_empty():
            # Remove and return the item from the front of the queue
            return self.queue.pop(0)

    def is_empty(self):
        # Check if the queue is empty
        return len(self.queue) == 0

    def size(self):
        # Get the current size of the queue
        return len(self.queue)


# Create a new queue
queue = Queue()

# Enqueue elements
queue.enqueue(1)
queue.enqueue(2)
queue.enqueue(3)
queue.enqueue(4)
queue.enqueue(5)

# Dequeue elements
item = queue.dequeue()
print("Dequeued item:", item)

item = queue.dequeue()
print("Dequeued item:", item)

# Check if the queue is empty
print("Is the queue empty?", queue.is_empty())

# Get the size of the queue
print("Queue size:", queue.size())

In this example, we create a Queue class with the following methods:

  • enqueue: Adds an item to the rear of the queue.
  • dequeue: Removes and returns the item from the front of the queue.
  • is_empty: Checks if the queue is empty.
  • size: Returns the current size of the queue.
  1. Solving a LeetCode Problem: "Implement Queue using Stacks" Now, let's solve a question from LeetCode that involves implementing a queue using stacks.

Problem: "Implement Queue using Stacks" You can find the problem details and constraints here: https://leetcode.com/problems/implement-queue-using-stacks/

Here's the complete code for solving the "Implement Queue using Stacks" problem:


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

    def push(self, x):
        # Move all elements from stack1 to stack2
        while self.stack1:
            self.stack2.append(self.stack1.pop())

        # Add the new element to stack1
        self.stack1.append(x)

        # Move all elements back to stack1
        while self.stack2:
            self.stack1.append(self.stack2.pop())

    def pop(self):
        if self.stack1:
            # Pop the top element from stack1
            return self.stack1.pop()

    def peek(self):
        if self.stack1:
            # Get the top element from stack1
            return self.stack1[-1]

    def empty(self):
        # Check if both stacks are empty
        return not self.stack1 and not self.stack2


# Example usage
queue = MyQueue()

queue.push(1)
queue.push(2)
queue.push(3)

print(queue.peek())  # Output: 1

print(queue.pop())  # Output: 1

print(queue.empty())  # Output: False

The MyQueue class implements a queue using two stacks (stack1 and stack2). The push operation is implemented by transferring all elements from stack1 to stack2, adding the new element to stack1, and then transferring all elements back to stack1. This ensures that the oldest element is always at the top of stack1, simulating the behavior of a queue.

  1. Time and Space Complexity Analysis: The time complexity of the push operation is O(N) in the worst case, where N is the number of elements in the queue. This is because we need to transfer all elements between the two stacks.

The time complexity of the pop, peek, and empty operations is O(1) since we directly access the top element or check if the stacks are empty.

The space complexity is O(N) since we use two stacks to store the elements of the queue.

For the "Implement Queue using Stacks" problem, the time complexity is O(N) for the push operation and O(1) for the other operations. The space complexity is O(N) due to the two stacks.

By understanding the concept of a queue and implementing the necessary methods, we can efficiently solve problems that require the queue's functionality, such as processing elements in a specific order.

 

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