What is the time complexity of Design Front Middle Back Queue?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Design Front Middle Back Queue?

This problem uses Array, Linked List, Design, Queue, Data Stream. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Design Front Middle Back Queue?

Insight 1: The key is to efficiently handle insertions and deletions from the front, middle, and back. Using a single data structure like an array or linked list would make operations like pushMiddle or popMiddle potentially O(n) because you need to shift elements. Insight 2: Using two deques (double-ended queues) can efficiently solve this problem. One deque (`left`) stores the left half of the queue and the other (`right`) stores the right half. This allows for O(1) insertions and deletions from both ends.

Which companies ask LeetCode 1670. Design Front Middle Back Queue in interviews?

This problem is frequently asked at Citadel.

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Design Front Middle Back Queue - LeetCode 1670 Solution

Design Front Middle Back Queue - Complete Solution Guide

Design Front Middle Back Queue is LeetCode problem 1670, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design a queue that supports push and pop operations in the front, middle, and back. Implement the FrontMiddleBack class: FrontMiddleBack() Initializes the queue. void pushFront(int val) Adds val to the front of the queue. void pushMiddle(int val) Adds val to the middle of the queue. void pushBack(int val) Adds val to the back of the queue. int popFront() Removes the front element of the queue and returns it. If the queue is empty, return -1 . int popMiddle() Removes the middle element of the qu

Detailed Explanation

The problem requires designing a queue-like data structure that supports push and pop operations from the front, middle, and back. The `FrontMiddleBackQueue` class must implement the following methods: * `pushFront(int val)`: Adds `val` to the front of the queue. * `pushMiddle(int val)`: Adds `val` to the middle of the queue. If the queue has an odd number of elements, the new element is inserted at the index (n-1)/2. If the queue has an even number of elements, the new element is inserted at index n/2. The example given clarifies the middle is always inserted towards the front when there are two possible middle indices. * `pushBack(int val)`: Adds `val` to the back of the queue. * `popFront()`: Removes and returns the element at the front of the queue. Returns -1 if the queue is empty. * `popMiddle()`: Removes and returns the element at the middle of the queue. Returns -1 if the queue is empty. If there are two middle elements (even size queue), remove the leftmost middle element (index n/2 - 1). * `popBack()`: Removes and returns the element at the back of the queue. Returns -1 if the queue is empty. The problem emphasizes the importance of handling the 'middle' element correctly when the queue size is even.

Solution Approach

The solution utilizes two deques: `left` and `right`. The `left` deque stores the left half of the queue, including the middle element when the queue size is odd. The `right` deque stores the right half of the queue. A `rebalance` function is used after each push/pop operation to ensure the deques are properly balanced. This approach allows efficient O(1) operations for pushing and popping from front, middle and back.

Step-by-Step Algorithm

Step 1: Initialize two deques, `left` and `right`, in the constructor.
Step 2: Implement `pushFront(val)`: Add `val` to the front of the `left` deque and call the `rebalance()` function.
Step 3: Implement `pushMiddle(val)`: If `len(left) > len(right)`, move the last element of `left` to the front of `right`. Then add `val` to the back of `left`.
Step 4: Implement `pushBack(val)`: Add `val` to the back of the `right` deque and call the `rebalance()` function.
Step 5: Implement `popFront()`: If `left` is empty, return -1. Otherwise, remove and return the front element of `left` and call the `rebalance()` function.
Step 6: Implement `popMiddle()`: If `left` is empty, return -1. Otherwise, remove and return the last element of `left` and call the `rebalance()` function.
Step 7: Implement `popBack()`: If `left` is empty, return -1. Otherwise, if `right` is not empty, remove and return the last element of `right`. If `right` is empty, remove and return the last element of `left`. Then call the `rebalance()` function.
Step 8: Implement `rebalance()`: If `len(left) > len(right) + 1`, move the last element of `left` to the front of `right`. If `len(left) < len(right)`, move the front element of `right` to the back of `left`.

Key Insights

Insight 1: The key is to efficiently handle insertions and deletions from the front, middle, and back. Using a single data structure like an array or linked list would make operations like pushMiddle or popMiddle potentially O(n) because you need to shift elements.
Insight 2: Using two deques (double-ended queues) can efficiently solve this problem. One deque (`left`) stores the left half of the queue and the other (`right`) stores the right half. This allows for O(1) insertions and deletions from both ends.
Insight 3: Maintaining balance between the two deques is crucial to ensure that the middle element can be accessed in O(1) time. The rebalancing operation ensures that `abs(len(left) - len(right))` is at most 1 at all times.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Array, Linked List, Design, Queue, Data Stream.

Companies

Asked at: Citadel.

Design Front Middle Back Queue | LeetCode 1670

Medium

Array Linked List Design Queue Data Stream

Time: O(1)Space: O(n)

Solution Code

Available in:

import java.util.Deque;
import java.util.LinkedList;

class FrontMiddleBackQueue {

    private Deque<Integer> left;
    private Deque<Integer> right;

    public FrontMiddleBackQueue() {
        left = new LinkedList<>();
        right = new LinkedList<>();
    }

    private void rebalance() {
        if (left.size() > right.size() + 1) {
            right.addFirst(left.removeLast());
        } else if (left.size() < right.size()) {
            left.addLast(right.removeFirst());
        }
    }

    public void pushFront(int val) {
        left.addFirst(val);
        rebalance();
    }

    public void pushMiddle(int val) {
        if (left.size() > right.size()) {
            right.addFirst(left.removeLast());
        }
        left.addLast(val);
    }

    public void pushBack(int val) {
        right.addLast(val);
        rebalance();
    }

    public int popFront() {
        if (left.isEmpty()) {
            return -1;
        }
        int val = left.removeFirst();
        rebalance();
        return val;
    }

    public int popMiddle() {
        if (left.isEmpty()) {
            return -1;
        }
        int val = left.removeLast();
        rebalance();
        return val;
    }

    public int popBack() {
        if (left.isEmpty()) {
            return -1;
        }

        int val;
        if (!right.isEmpty()) {
            val = right.removeLast();
        } else {
            val = left.removeLast();
        }

        rebalance();
        return val;
    }
}

/**
 * Your FrontMiddleBackQueue object will be instantiated and called as such:
 * FrontMiddleBackQueue obj = new FrontMiddleBackQueue();
 * obj.pushFront(val);
 * obj.pushMiddle(val);
 * obj.pushBack(val);
 * int param_4 = obj.popFront();
 * int param_5 = obj.popMiddle();
 * int param_6 = obj.popBack();
 */

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Problem Info

Description

Problem 1670: Design Front Middle Back Queue - Detailed explanation and solution approaches.

Topics

Array Linked List Design Queue Data Stream

Companies

Citadel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1670

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize two deques, `left` and `right`, in the constructor.
2
Step 2: Implement `pushFront(val)`: Add `val` to the front of the `left` deque and call the `rebalance()` function.
3
Step 3: Implement `pushMiddle(val)`: If `len(left) > len(right)`, move the last element of `left` to the front of `right`. Then add `val` to the back of `left`.
4
Step 4: Implement `pushBack(val)`: Add `val` to the back of the `right` deque and call the `rebalance()` function.
5
Step 5: Implement `popFront()`: If `left` is empty, return -1. Otherwise, remove and return the front element of `left` and call the `rebalance()` function.
6
Step 6: Implement `popMiddle()`: If `left` is empty, return -1. Otherwise, remove and return the last element of `left` and call the `rebalance()` function.
7
Step 7: Implement `popBack()`: If `left` is empty, return -1. Otherwise, if `right` is not empty, remove and return the last element of `right`. If `right` is empty, remove and return the last element of `left`. Then call the `rebalance()` function.
8
Step 8: Implement `rebalance()`: If `len(left) > len(right) + 1`, move the last element of `left` to the front of `right`. If `len(left) < len(right)`, move the front element of `right` to the back of `left`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Deque operations like `appendleft`, `append`, `popleft`, and `pop` are all O(1). The `rebalance` function also takes O(1) time because it only involves moving a single element between the deques. The space complexity is O(n) because, in the worst case, all elements will be stored in the deques.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Doubly Linked List

A doubly linked list could be used. You'd need to maintain a pointer to the middle element to efficiently insert/delete there. The head and tail pointers allow O(1) insertion/deletion at the front and back.

Time: O(1) for all operations, assuming the middle pointer is maintained correctly.Space: O(n)

Trade-offs: Doubly Linked List might be slightly more space-efficient in some implementations but requires manual memory management in languages like C/C++. Maintaining the middle pointer adds complexity.

✨

Key Insights

1Insight 1: The key is to efficiently handle insertions and deletions from the front, middle, and back. Using a single data structure like an array or linked list would make operations like pushMiddle or popMiddle potentially O(n) because you need to shift elements.
2Insight 2: Using two deques (double-ended queues) can efficiently solve this problem. One deque (`left`) stores the left half of the queue and the other (`right`) stores the right half. This allows for O(1) insertions and deletions from both ends.
3Insight 3: Maintaining balance between the two deques is crucial to ensure that the middle element can be accessed in O(1) time. The rebalancing operation ensures that `abs(len(left) - len(right))` is at most 1 at all times.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Queue is empty when popFront, popMiddle, or popBack are called. The solution correctly returns -1 in this case.
Edge case 2: Pushing to the middle with an even or odd number of elements. The algorithm handles this correctly by checking the lengths of the left and right deques before pushing.
Edge case 3: Popping from the back when the right deque is empty. The algorithm handles this by popping from the left deque if the right deque is empty.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to rebalance the deques after push/pop operations. This can lead to incorrect results when accessing the middle element.

✗Common mistake 2: Incorrectly calculating the middle index. Pay careful attention to whether the queue size is even or odd and adjust the insertion/deletion accordingly.

✗Common mistake 3: Not handling the edge case where the queue is empty when attempting to pop an element. This will cause runtime errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Implementing a text editor with undo/redo functionality. You can insert text at the beginning or end and revert to a previous state by undoing (popping from the front) or redoing (pushing to the back). Pushing to the middle can be used for merging changes from other users.

🏦Real-world scenario 2: Managing a buffer in a video streaming application. The front of the queue represents the frames that have been played, the back represents the frames that are ready to be played, and the middle can be used to seek to a specific frame.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Design Front Middle Back Queue"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether linked list techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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