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Solving 5 common coding interview problems with queues

Q: What are the basic operations of a queue?

The main operations of a queue are: - **Enqueue:** Adds an item to the end of the queue. - **Dequeue:** Removes an item from the front of the queue and returns it. - **Peek:** Shows the item at the front of the queue without removing it. - **Is Empty:** Checks if the queue has no items.

14 min read

As an interviewer at FAANG companies and Educative, I’ve seen how choosing the right data structure can be the difference between an optimal solution and a missed opportunity, especially when time is of the essence. Understanding data structures deeply is crucial for coding interviews, as your choice directly affects time and space complexity.

One essential data structure is the queue, which is indispensable for real-world problems and interview challenges where maintaining order is key. A queue operates on a first-in, first-out (FIFO) principle, where items are processed in the order they arrive—just like a line at a ticket counter.

In this blog, I will cover the following:

How data structures are essential to design optimized algorithms
The fundamental concept and basic operations of queues.
Different variations of queues:
- Circular Queue
- Priority Queue
- Deque
- Concurrent Queue
- Blocking Queue
Identifying problems that can be solved using queues
5 coding problems & solutions using queues:
- Level order traversal of binary tree
- Find the maximum in sliding window
- Binary tree zigzag level order traversal
- Course schedule II
- Number of islands

Basic operations of a queue#

Like any other data structure, a queue offers some basic operations for data manipulation. We can insert (enqueue) and remove (dequeue) data to and from a queue. We can also check whether the queue’s current state is empty. In the primary variation of a queue, an item can only be enqueued at the rear, and an item can only be removed from the front. On the other hand, the queue does not allow manipulation of the data from the center.

The following are the basic operations a queue offers:

Enqueue: Adds an element at the queue end.
Dequeue: Removes an element from the queue front and returns it.
Peek: Returns the element from the queue front without removing it.
Is Empty: Check if the queue does not contain any item.

Let’s see the queue implementation, which includes different methods to perform the aforementioned operations.

# Node class represents an individual element in the linked list.
class Node:
    def __init__(self, value):
        self.value = value  # Value of the node
        self.next = None    # Pointer to the next node
# Queue class implements queue operations using a linked list.
class Queue:
    def __init__(self):
        self.front = self.rear = None  # Pointers to the front and rear of the queue
    # Method to check if the queue is empty
    def is_empty(self):
        return self.front is None
    # Method to add a new element to the rear of the queue
    def enqueue(self, value):
        new_node = Node(value)  # Create a new node with the given value
        if self.rear is None:  # If the queue is empty, both front and rear will point to the new node
            self.front = self.rear = new_node
            return
        self.rear.next = new_node  # Link the new node to the end of the queue
        self.rear = new_node  # Update the rear pointer to the new node
    # Method to remove an element from the front of the queue
    def dequeue(self):
        if self.is_empty():  # If the queue is empty, raise an exception
            raise IndexError("Dequeue from an empty queue")
        value = self.front.value  # Store the value to return
        self.front = self.front.next  # Move the front pointer to the next node
        if self.front is None:  # If the queue is now empty, set rear to None
            self.rear = None
        return value  # Return the dequeued value
    # Method to get the value at the front of the queue without removing it
    def peek(self):
        if self.is_empty():  # If the queue is empty, raise an exception
            raise IndexError("Peek from an empty queue")
        return self.front.value  # Return the value at the front
# Main code for testing the Queue class
if __name__ == "__main__":
    queue = Queue()  # Create a new queue
    queue.enqueue(10)  # Add 10 to the queue
    queue.enqueue(20)  # Add 20 to the queue
    print(queue.peek())    # Output: 10
    print(queue.dequeue()) # Output: 10
    print(queue.dequeue()) # Output: 20
    print(queue.is_empty()) # Output: True

Different variations of the queue#

The basic concept of a queue is to enqueue new elements at the rear and dequeue existing elements from the front. Several variations of the queue are also designed to address specific needs and optimize performance for different scenarios. Let’s discuss the various types of queues and highlight their unique characteristics.

Circular Queue#

A specialized queue data structure where the last element connects back to the first, forming a circular arrangement. This design optimizes memory usage, especially for fixed-size queues. Unlike a standard queue, which requires shifting elements to reuse empty spaces, a circular queue allows continuous usage of memory locations without overhead. The empty spaces appear when we dequeue an element from the front of the queue. Instead of wasting the space, a circular queue connects its rear back to the front to use that freed-up space from enqueuing new elements. Task Scheduling and Memory allocation are examples of using circular queues in practical applications. Task scheduling in operating systems requires processes to be scheduled. Round Robin is one such algorithm where a circular queue of processes is maintained to assign CPU time. For managing data packets in network data management, a circular queue ensures that the packets are processed in the order they arrive and reuse the space efficiently.

Deque#

A deque (pronounced “deck”), also known as a double-ended queue, is a variation of the queue that allows elements to be added or removed from both ends. This means we can add or remove an element on the queue’s front or rear end. A deque can be further classified into input-restricted dequeA type of deque where elements can only be removed from both ends, but insertion is only allowed at one end. and output-restricted dequeA type of deque where elements can only be added from both ends, but removal is restricted to one end.. The most common application of the deque is checking if a string is palindrome. This is done by comparing string characters from both ends of the deque. Deques are also used to manage the browser history. This is accomplished by adding new pages to the deque and navigating backward and forward by removing from either of the ends.

How to identify problems that can be solved using queues#

Learning to quickly identify queue-appropriate problems is crucial for tech interviews. We can solve a problem using a queue if the problem has one or more of the following indications:

First-in-first-out (FIFO) order of the elements:
- If we are solving a problem that requires processing the elements in the sequence of their arrival, the queue would be the best data structure.
  - Using the queue, we will ensure that we follow the exact requirements of the problem by processing the items that arrive first and putting the new elements at the end of the queue.
  - The common problems include task scheduling, where we want to execute the tasks in the order they arrive and print jobs to print the documents as they are enqueuing.
Level order traversal:
- A queue can solve problems requiring traversing a data structure’s elements level by level.
- Using a queue helps manage the nodes at each level. This way, we can ensure we processed all nodes at one level before moving on to the next level.
- A breadth-first traversal of the trees and graphs is a common problem that we can solve using level order traversal.
Producer-consumer problem:
- When solving a problem where some producers generate data and one or more consumers process that data, a queue can store the data generated by the producer in a buffer for consumers to use.
- Queues ensures smooth data flow between the producer and consumer.
- The example includes the logging system, where the queue will store the system logs as they arrive before storing them on the disk. Data streaming is another example where a queue helps store a stream’s data before presenting it to the user.
Sliding window problems:
- When solving problems that require tracking a subset of elements within a fixed-size window moving through a dataset, we often use queues.
- A queue helps manage the elements in the current window, making it easy to update the window as new elements enter and old ones leave.
- A typical example of identifying the sliding window pattern is finding the maximum or minimum value in a sliding window using the queue to record the elements within the current window.

Solving 5 queue coding interview questions#

Let’s talk about the coding problems that can be solved using the queue or one of its variations.

1. Level order traversal of binary tree#

Display the nodes’ values of a binary tree using a level order traversal.

Solution#

The optimized solution to this problem is to use a deque to traverse the tree’s nodes level-by-level efficiently.

Create the following two lists and deques:

A deque for traversing the tree level by level
An empty list for nodes at the current level
A list of nodes from all levels

Enqueue the root node into the deque and loop until the deque is empty. For each node, dequeue it, add it to the current level list, and enqueue its children if they exist. After processing all nodes at the current level, join the current-level nodes into a comma-separated string and add it to the result list. Finally, all level strings from the result list are joined into a single string with colons as separators and printed, representing the level order traversal of the binary tree.

The time complexity of this solution is linear, that is, $O(n)$ , because every node is visited only once. The space complexity of this solution is also linear, $O(n)$ , because the queue will keep $⌈n/2⌉$ nodes at maximum (at the level of leaf nodes) at any stage.

main.py

TreeNode.py

BinaryTree.py

from collections import deque
from BinaryTree import *
from TreeNode import *
# Function to perform level order traversal of a binary tree
def level_order_traversal(root):
    result = []  # Initialize an empty list to store the result
    if not root:
        result = "None"
        return result
    current_queue = deque()  # Initialize a deque for level order traversal
    current_queue.append(root)  # Start with the root node
    while current_queue:
        level_size = len(current_queue)  # Number of nodes at the current level
        level_nodes = []  # List to store nodes of the current level
        for _ in range(level_size):
            temp = current_queue.popleft()  # Remove the front node from the queue
            level_nodes.append(str(temp.data))  # Append the node's data to the level list
            # Add the left child to the queue if it exists
            if temp.left:
                current_queue.append(temp.left)
            # Add the right child to the queue if it exists
            if temp.right:
                current_queue.append(temp.right)
        # Append the current level's nodes to the result list
        result.append(", ".join(level_nodes))
    # Join all levels with " : " separator and return as a string
    return " : ".join(result)
# Main function to run test cases
def main():
    test_cases_roots = []  # List to store roots of test binary trees
    # Create the first binary tree
    input1 = [
        TreeNode(100),
        TreeNode(50),
        TreeNode(200),
        TreeNode(25),
        TreeNode(75),
        TreeNode(350)
    ]
    tree1 = BinaryTree(input1)  # Initialize BinaryTree with input1
    test_cases_roots.append(tree1.root)  # Append the root of tree1 to test_cases_roots
    # Create the second binary tree
    input2 = [
        TreeNode(25),
        TreeNode(50),
        None,
        TreeNode(100),
        TreeNode(200),
        TreeNode(350)
    ]
    tree2 = BinaryTree(input2)  # Initialize BinaryTree with input2
    test_cases_roots.append(tree2.root)  # Append the root of tree2 to test_cases_roots
    # Create the third binary tree
    input3 = [
        TreeNode(350),
        None,
        TreeNode(100),
        None,
        TreeNode(50),
        TreeNode(25)
    ]
    tree3 = BinaryTree(input3)  # Initialize BinaryTree with input3
    test_cases_roots.append(tree3.root)  # Append the root of tree3 to test_cases_roots
    # Create the fourth binary tree with a single node
    tree4 = BinaryTree([TreeNode(100)])
    test_cases_roots.append(tree4.root)  # Append the root of tree4 to test_cases_roots
    # Append None to represent an empty tree
    test_cases_roots.append(None)
    # Iterate over each test case and print the level order traversal
    for i in range(len(test_cases_roots)):
        if i > 0:
            print()  # Print a newline for separation
        print(i + 1, ".\tBinary Tree")
        display_tree(test_cases_roots[i])  # Display the binary tree structure
        print("\n\tLevel order traversal: ")
        print("\t", level_order_traversal(test_cases_roots[i]))  # Print level order traversal
        print("\n" + '-' * 100)  # Print a separator line
# Run the main function if the script is executed
if __name__ == '__main__':
    main()

Level Order Traversal of Binary Tree

2. Find the maximum in sliding window#

Given an integer list, find the maximum values in all contiguous subarrays of a specific size.

Solution#

The optimized solution to this problem is using a sliding window and a queue. First, if the input list has only one element, return it immediately since no further processing is needed. Next, use a deque for the first $w$ elements to store the indices of potential maximum values. For each element, clean up the deque by removing indices of elements that are less than or equal to the current element’s value, then add the current element’s index to the deque. Once the first $w$ elements are processed, add the value at the front of the deque to the output list as the maximum for the first window.

Then, move through the rest of the input list, repeating the cleanup and append steps for each element. Before adding the current element’s index to the deque during each iteration, remove the first index if it’s no longer within the current window. Finally, return the list that contains the maximum values for each window.

When the window initially moves forward, and if the new element is larger than all other elements in the deque, the pop operation must be performed $w$ times. The pop operation is executed only once, as the deque will only contain a single element from that point onward. All these steps are calculated as $O(w+n−w)$ , that is, $O(n)$ . The space complexity is $O(w)$ , where $w$ is the window size.

from collections import deque
# function to clean up the deque
def clean_up(i, current_window, nums):
    while current_window and nums[i] >= nums[current_window[-1]]:
        current_window.pop()
# function to find the maximum in all possible windows
def find_max_sliding_window(nums, w):
    if len(nums) == 1:
        return nums
    output = []
    current_window = deque()
    for i in range(w):
        clean_up(i, current_window, nums)
        current_window.append(i)
    output.append(nums[current_window[0]])
    for i in range(w, len(nums)):
        clean_up(i, current_window, nums)
        if current_window and current_window[0] <= (i - w):
            current_window.popleft()
        current_window.append(i)
        output.append(nums[current_window[0]])
    return output
# driver code
def main():
    window_sizes = [3, 3, 3, 3, 2, 4, 3, 2, 3, 6]
    nums_list = [
        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
        [10, 9, 8, 7, 6, 5, 4, 3, 2, 1],
        [10, 10, 10, 10, 10, 10, 10, 10, 10, 10],
        [1, 5, 8, 10, 10, 10, 12, 14, 15, 19, 19, 19, 17, 14, 13, 12, 12, 12, 14, 18, 22, 26, 26, 26, 28, 29, 30],
        [10, 6, 9, -3, 23, -1, 34, 56, 67, -1, -4, -8, -2, 9, 10, 34, 67],
        [4, 5, 6, 1, 2, 3],
        [9, 5, 3, 1, 6, 3],
        [2, 4, 6, 8, 10, 12, 14, 16],
        [-1, -1, -2, -4, -6, -7],
        [4, 4, 4, 4, 4, 4]
    ]
    for i in range(len(nums_list)):
        print(f"{i + 1}.\tInput list:\t{nums_list[i]}")
        print(f"\tWindow size:\t{window_sizes[i]}")
        output = find_max_sliding_window(nums_list[i], window_sizes[i])
        print(f"\n\tMaximum in each sliding window:\t{output}")
        print("-"*100)
if __name__ == '__main__':
    main()

Find Maximum in Sliding Window

3. Binary tree zigzag level order traversal#

Given a binary tree, return its zigzag level order traversal. In this traversal, nodes are traversed from left to right at one level, then from right to left at the next level, and this alternating pattern continues, reversing the direction at each subsequent level.

Solution#

The optimized solution to this problem is to use the deque to track all the nodes at a level. We will use each level’s first-in-first-out (FIFO) and last-in-first-out (LIFO) schemes to perform the traversal in a zigzag level order. We append new elements to the deque’s tail to ensure FIFO ordering. Likewise, we append the new elements to the head of the deque to ensure LIFO ordering.

We will start by adding the tree’s root to a deque for level order traversal and set a reverse flag to manage alternation between FIFO and LIFO schemes. The reverse flag starts as false and toggles at each level. When it is false, pop nodes from the front (left) of the deque and push their children to the back (right), ensuring right-to-left traversal. When it is true, pop nodes from the back (right) and push their children to the front (left), ensuring left-to-right traversal.

This solution’s time complexity is $O(n)$ , where $n$ represents the number of nodes in the tree. Its space complexity will also be $O(n)$ , where $n$ is the space occupied by the deque.

main.py

BinaryTree.py

TreeNode.py

from BinaryTree import *
from TreeNode import *
from collections import deque
def zigzag_level_order(root):
    # If the root is NULL, return an empty list
    if root is None:
        return []
    # Creating an empty list to store the results
    results = []
    # Creating a deque with the root node as the only element
    dq = deque([root])
    # Creating a flag to indicate the direction of the traversal
    reverse = False
    # Traverse the tree
    while len(dq):
        # Getting the size of the current level
        size = len(dq)
        # Insert an empty list at the end of the 'results' list
        results.insert(len(results), [])
        # Traverse the nodes in the current level
        for i in range(size):
            # Check the direction of the traversal
            if not reverse:
                # If the direction is left-to-right, pop a node from the 
                # start (left) of the deque and add it to the current level
                node = dq.popleft()
                results[len(results) - 1].append(node.data)
                # Add the left and right child nodes 
                # of the current node to the deque
                if node.left:
                    dq.append(node.left)
                if node.right:
                    dq.append(node.right)
            else:
                # If the direction is right-to-left, pop the a node from the
                # back (right) of the deque and add it to the current level
                node = dq.pop()
                results[len(results) - 1].append(node.data)
                # Add the right and left child nodes 
                # of the current node to the deque
                if node.right:
                    dq.appendleft(node.right)
                if node.left:
                    dq.appendleft(node.left) 
        # Reverse the direction of traversal for the next level
        reverse = not reverse
    # Return the results
    return results
# Driver code
def main():
    # Creating a binary tree
    input1 = [TreeNode(100), TreeNode(50), TreeNode(200), TreeNode(25), TreeNode(75), TreeNode(350)]
    tree1 = BinaryTree(input1)
    # Creating a right degenerate binary tree
    input2 = [TreeNode(25), TreeNode(50), TreeNode(75), TreeNode(100), TreeNode(200), TreeNode(350)]
    tree2 = BinaryTree(input2)
    # Creating a left degenerate binary tree
    input3 = [TreeNode(350), TreeNode(200), TreeNode(100), TreeNode(75), TreeNode(50), TreeNode(25)]
    tree3 = BinaryTree(input3)
    # Creating a single node binary tree
    tree4 = BinaryTree([TreeNode(100)])
    roots = [tree1.root, tree2.root, tree3.root, tree4.root, None]
    for i in range(len(roots)):
        print(i+1, ".\tBinary Tree:", sep = "")
        display_tree(roots[i])
        print("\n\tThe zigzag level order traversal is:", zigzag_level_order(roots[i]))
        print("\n", "-"*100, "\n", sep = "")
if __name__ == '__main__':
    main()

Binary Tree Zigzag Level Order Traversal

4. Course schedule #

Consider a scenario with a set of courses labeled sequentially, some of which have prerequisites. The prerequisites are given as pairs, indicating that one course must be completed before the other. The task is to determine a valid order for students to complete all the courses.

Solution#

The optimized solution to this problem is to use a queue and topological sort to find the linear ordering of elements.

The graph is stored using adjacency lists managed through a hash map, where each parent vertex number is a key linked to a list of child vertex numbers. We use another hash map to track the in-degrees of each vertex, indicating the count of incoming edges. Vertices with an in-degree of 0 are considered sources. The graph is built from input data while updating the in-degrees hash map. Sources are added to a queue for processing. For each source in the queue, we add it to the sorted list, retrieve its children, decrease each child’s in-degree by 1, and if a child’s in-degree becomes 0, add it to the source queue. We will repeat this process until the queue is empty.

The time complexity of this solution is $O(v+e)$ , where v and e represent the number of vertices and edges in the graph, respectively. The space complexity is also $O(v+e)$ because all edges of each vertex are stored in an adjacency list.

from collections import deque
def find_order(n, prerequisites):
    sorted_order = []
    # if n is smaller than or equal to zero we will return the empty array
    if n <= 0:
        return sorted_order
    # Store the count of incoming prerequisites in a hashmap
    in_degree = {i: 0 for i in range(n)}  
    # a. Initialize the graph
    graph = {i: [] for i in range(n)}  # adjacency list graph
    # b. Build the graph
    for prerequisite in prerequisites:
        parent, child = prerequisite[1], prerequisite[0]
        graph[parent].append(child)  # add the child to its parent's list
        in_degree[child] += 1  # increment child's in_degree
    # c. Find all sources i.e., all nodes with 0 in-degrees
    sources = deque()
    # traverse in in_degree using key
    for key in in_degree:
        # if in_degree[key] is 0 append the key in the deque sources
        if in_degree[key] == 0:
            sources.append(key)
    # d. For each source, add it to the sorted_order and subtract one from 
    # all of its children's in-degrees. If a child's in-degree becomes zero, 
    # add it to the sources queue
    while sources:
        # pop an element from the start of the sources and store 
        # the element in vertex
        vertex = sources.popleft()
        # append the vertex at the end of the sorted_order
        sorted_order.append(vertex)
        # traverse in graph[vertex] using child
        # get the node's children to decrement 
        # their in-degrees
        for child in graph[vertex]:  
            in_degree[child] -= 1
            # if in_degree[child] is 0 append the child in the deque sources
            if in_degree[child] == 0:
                sources.append(child)
    # topological sort is not possible as the graph has a cycle
    if len(sorted_order) != n:
        return []
    return sorted_order
# Driver code
def main():
    n = [4, 5, 0, 4, 7]
    prerequisites = [
      [[1, 0], [2, 0], [3, 1], [3, 2]], 
      [[1, 0], [2, 0], [3, 1], [4, 3]], 
      [[1, 0]], [[1, 0], [2, 0], [3, 1], [3, 2]], 
      [[1, 0], [0, 3], [0, 2], [3, 2], [2, 5], [4, 5], [5, 6], [2, 4]]]
    for i in range(len(n)):
        print(i + 1, ".\tPrerequisites: ", prerequisites[i], sep="")
        print("\tTotal number of courses, n:", n[i])
        print("\tValid courses order:", find_order(n[i], prerequisites[i]))
        print("-"*100)
if __name__ == '__main__':
    main()

Course Schedule II

5. Number of islands#

Imagine a scenario with an $m∗n$ 2D grid filled with binary numbers, where $0$ represents water and $1$ represents land. If any cells with $1$ are connected horizontally or vertically (but not diagonally), they form an island. The task is to determine the total number of islands in the grid.

Solution#

The optimized solution to this problem is to use a queue and explore the island with the help of Breadth-First-Search (BFS).

To find the number of islands in a grid, initialize a counter to keep track of the number of islands found. Iterate through each cell in the grid and use a queue to explore all parts of an island, starting from the first land cell (the cell with value $1$ ). Mark this cell as visited by setting it to $0$ . Add the starting cell to the queue, and while the queue is not empty, pop the front cell and check all four possible directions. If a neighboring cell contains $1$ , add it to the queue by marking it as visited. After completely exploring an island using BFS, increment the island count. At the end, return the value of the counter.

This solution’s time complexity is $O(m∗n)$ , where $m$ and $n$ represent rows and columns, respectively. The problem’s space complexity is $O(min(m, n))$ because the queue may grow up to $m$ or $n$ , whichever is smaller in value.

def numIslands(grid):
    if not grid:
        return 0
    # Initialize the count of islands
    island_count = 0
    # Define the DFS function to explore the island
    def dfs(i, j):
        # Check if the current position is out of the grid or is water ('0')
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == '0':
            return
        # Mark the current land as visited by setting it to '0'
        grid[i][j] = '0'
        # Visit all 4 neighboring cells (up, down, left, right)
        dfs(i + 1, j)  # down
        dfs(i - 1, j)  # up
        dfs(i, j + 1)  # right
        dfs(i, j - 1)  # left
    # Iterate through each cell in the grid
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            # If the cell is land ('1'), it's part of an island
            if grid[i][j] == '1':
                # Increment the island count
                island_count += 1
                # Start a DFS to mark all parts of this island
                dfs(i, j)
    # Return the total number of islands found
    return island_count
# Input:
grid = [
    ["1", "1", "0", "0", "0"],
    ["1", "1", "0", "0", "0"],
    ["0", "0", "1", "0", "0"],
    ["0", "0", "0", "1", "1"]
]
# Printing the output
print(numIslands(grid))

Course Schedule II

From data structures to patterns#

Mastering the queue data structure is a critical step toward solving a wide range of coding interview problems, from breadth-first search in trees to task scheduling in operating systems. Queue is often the go-to whenever order and sequence matter. By understanding how and when to use queues, you can tackle problems more quickly in the coding interview and on the job.

That said, mastering individual data structures like queues is not enough to succeed. To truly excel in coding interviews, your best bet is to focus on identifying common patterns that underly all coding problems. Coding patterns will help you identify the optimal solution by recognizing a problem's underlying structure—saving precious time during interviews. While it's essential to keep refining your understanding of data structures, you should start thinking in patterns to truly prepare for the unknown in the interview.

You can learn about 26 coding interview patterns that helps solve hundreds of problems (including the 5 we solved today) with Educative's Grokking Coding Interview Patterns series, check them out below.

Frequently Asked Questions

What is a queue, and how does it differ from other data structures?

A queue is a basic data structure that follows the first in, first out (FIFO) principle, meaning that elements are processed in the order they arrive. Unlike arrays or linked lists, a queue only allows you to add items at the back and remove them from the front. This ensures that the oldest elements are handled first.

What are the basic operations of a queue?

The main operations of a queue are:

Enqueue: Adds an item to the end of the queue.
Dequeue: Removes an item from the front of the queue and returns it.
Peek: Shows the item at the front of the queue without removing it.
Is Empty: Checks if the queue has no items.

What is a circular queue, and when is it useful?

A circular queue is one where the end connects to the beginning, forming a loop. This design helps make better use of memory by reusing space efficiently. It’s especially useful in scenarios like task scheduling or resource management, where the queue needs to be continuously reused.

How does a priority queue differ from a standard queue?

In a priority queue, each item has a priority level. Items with higher priority are processed before those with lower priority, regardless of the order in which they were added. In contrast, a standard queue processes items strictly in the order they arrive.

What problem-solving scenarios are best suited for using a queue?

Queues are great for situations where you need to handle items in the order they come in, such as:

Traversing trees or graphs level by level.
Managing a sliding data window to keep track of elements within a fixed size.
Buffering data between producers and consumers, like in print jobs or task scheduling.

Written By:

Fahim ul Haq

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Solving 5 common coding interview problems with queues

Basic operations of a queue#

Different variations of the queue#

Circular Queue#

Priority Queue#

Deque#

Concurrent Queue#

Blocking Queue#

How to identify problems that can be solved using queues#

Solving 5 queue coding interview questions#

1. Level order traversal of binary tree#

Solution#

2. Find the maximum in sliding window#

Solution#

3. Binary tree zigzag level order traversal#

Solution#

4. Course schedule #

Solution#

5. Number of islands#

Solution#

From data structures to patterns#

Frequently Asked Questions

What is a queue, and how does it differ from other data structures?

What are the basic operations of a queue?

What is a circular queue, and when is it useful?

How does a priority queue differ from a standard queue?

What problem-solving scenarios are best suited for using a queue?