What is the time complexity of All O`one Data Structure?

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

What data structures are used to solve All O`one Data Structure?

This problem uses Hash Table, Linked List, Design, Doubly-Linked List. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving All O`one Data Structure?

Insight 1: Maintaining a doubly-linked list of buckets, where each bucket stores keys with the same count, allows for O(1) access to minimum and maximum counts. Insight 2: A hash map mapping keys to their respective nodes in the linked list enables O(1) increment and decrement operations.

Which companies ask LeetCode 432. All O`one Data Structure in interviews?

This problem is frequently asked at Airbnb, Atlassian, LinkedIn, Media.net, Nextdoor.

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All O`one Data Structure - LeetCode 432 Solution

All O`one Data Structure - Complete Solution Guide

All O`one Data Structure is LeetCode problem 432, a Hard 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 data structure to store the strings' count with the ability to return the strings with minimum and maximum counts. Implement the AllOne class: AllOne() Initializes the object of the data structure. inc(String key) Increments the count of the string key by 1 . If key does not exist in the data structure, insert it with count 1 . dec(String key) Decrements the count of the string key by 1 . If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed t

Detailed Explanation

The problem requires designing a data structure called `AllOne` that efficiently stores the counts of strings and allows retrieval of strings with the maximum and minimum counts in O(1) time. The data structure should support incrementing the count of a given string, decrementing the count of a string, and removing the string if its count becomes zero. The constraints limit the key length to 10 lowercase English letters, guarantee that `dec` is only called on existing keys, and restrict the total number of calls to the methods to 5 * 10^4.

Solution Approach

The solution utilizes a combination of a doubly-linked list and a hash map. The doubly-linked list stores nodes (buckets), each representing a specific count. Each node holds a set of keys sharing the same count. The hash map maps each key to its corresponding node in the doubly-linked list. Incrementing a key involves moving it to the next higher count node, creating a new node if necessary. Decrementing involves moving it to the next lower count node or removing the key and node if the count becomes zero. `getMaxKey()` and `getMinKey()` simply return a key from the tail and head of the linked list respectively, achieving O(1) time complexity.

Step-by-Step Algorithm

Step 1: Initialize the `AllOne` data structure with a doubly-linked list with head and tail dummy nodes, and an empty hash map `key_to_node`.
Step 2: Implement the `inc(key)` method: If the key exists in the `key_to_node` map, get its current node. Otherwise, set the current node to the head.
Step 3: Increment the count and move the key to the appropriate bucket. If a bucket with the incremented count does not exist, create one.
Step 4: Update the `key_to_node` map with the new node for the key.
Step 5: Implement the `dec(key)` method: Get the current node of the key from `key_to_node`. Decrement the count and move the key to the appropriate bucket. If the new count is 0, remove the key.
Step 6: If the count becomes zero, remove the entry from the keyToNode map. Otherwise update the mapping.
Step 7: Implement `getMaxKey()` by returning a key from the node before the tail. If the list is empty return an empty string.
Step 8: Implement `getMinKey()` by returning a key from the node after the head. If the list is empty return an empty string.

Key Insights

Insight 1: Maintaining a doubly-linked list of buckets, where each bucket stores keys with the same count, allows for O(1) access to minimum and maximum counts.
Insight 2: A hash map mapping keys to their respective nodes in the linked list enables O(1) increment and decrement operations.
Insight 3: Correctly handling edge cases such as an empty AllOne data structure or the first or last node being updated is crucial for correct functionality.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Linked List, Design, Doubly-Linked List.

Companies

Asked at: Airbnb, Atlassian, LinkedIn, Media.net, Nextdoor.

All O`one Data Structure | LeetCode 432

Hard

Hash Table Linked List Design Doubly-Linked List

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class AllOne {

    class Node {
        int count;
        Set<String> keys;
        Node prev;
        Node next;

        public Node() {
            this(0);
        }
        public Node(int count) {
            this.count = count;
            this.keys = new HashSet<>();
            this.prev = null;
            this.next = null;
        }
    }

    private Map<String, Node> keyToNode;
    private Node head;
    private Node tail;

    public AllOne() {
        keyToNode = new HashMap<>();
        head = new Node();
        tail = new Node();
        head.next = tail;
        tail.prev = head;
    }

    private void insertAfter(Node prevNode, Node newNode) {
        newNode.prev = prevNode;
        newNode.next = prevNode.next;
        prevNode.next.prev = newNode;
        prevNode.next = newNode;
    }

    private void removeNode(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }

    public void inc(String key) {
        Node currentNode;
        int newCount;
        if (!keyToNode.containsKey(key)) {
            currentNode = head;
            newCount = 1;
        } else {
            currentNode = keyToNode.get(key);
            currentNode.keys.remove(key);
            newCount = currentNode.count + 1;
        }

        Node targetNode = currentNode.next;
        if (targetNode.count != newCount) {
            targetNode = new Node(newCount);
            insertAfter(currentNode, targetNode);
        }

        targetNode.keys.add(key);
        keyToNode.put(key, targetNode);

        if (currentNode != head && currentNode.keys.isEmpty()) {
            removeNode(currentNode);
        }
    }

    public void dec(String key) {
        Node currentNode = keyToNode.get(key);
        currentNode.keys.remove(key);

        int currentCount = currentNode.count;
        if (currentCount > 1) {
            int newCount = currentCount - 1;
            Node targetNode = currentNode.prev;
            if (targetNode.count != newCount) {
                targetNode = new Node(newCount);
                insertAfter(currentNode.prev, targetNode);
            }
            targetNode.keys.add(key);
            keyToNode.put(key, targetNode);
        } else {
            keyToNode.remove(key);
        }

        if (currentNode.keys.isEmpty()) {
            removeNode(currentNode);
        }
    }

    public String getMaxKey() {
        if (tail.prev == head) {
            return "";
        }
        return tail.prev.keys.iterator().next();
    }

    public String getMinKey() {
        if (head.next == tail) {
            return "";
        }
        return head.next.keys.iterator().next();
    }
}

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Problem Info

Description

Problem 432: All O`one Data Structure - Detailed explanation and solution approaches.

Topics

Hash Table Linked List Design Doubly-Linked List

Companies

AirbnbAtlassianLinkedInMedia.netNextdoor

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:432

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 the `AllOne` data structure with a doubly-linked list with head and tail dummy nodes, and an empty hash map `key_to_node`.
2
Step 2: Implement the `inc(key)` method: If the key exists in the `key_to_node` map, get its current node. Otherwise, set the current node to the head.
3
Step 3: Increment the count and move the key to the appropriate bucket. If a bucket with the incremented count does not exist, create one.
4
Step 4: Update the `key_to_node` map with the new node for the key.
5
Step 5: Implement the `dec(key)` method: Get the current node of the key from `key_to_node`. Decrement the count and move the key to the appropriate bucket. If the new count is 0, remove the key.
6
Step 6: If the count becomes zero, remove the entry from the keyToNode map. Otherwise update the mapping.
7
Step 7: Implement `getMaxKey()` by returning a key from the node before the tail. If the list is empty return an empty string.
8
Step 8: Implement `getMinKey()` by returning a key from the node after the head. If the list is empty return an empty string.

💡

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?

The time complexity is O(1) due to the hash map providing O(1) lookup for keys, and the doubly linked list allowing O(1) insertion and deletion of nodes. The space complexity is O(n) because, in the worst case, all the input keys are unique, leading to a space proportional to the number of distinct keys.

📖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

Using a Sorted Map

Using a sorted map where the key is the count and the value is a set of strings. This makes getting the min and max keys relatively easy. However incrementing and decrementing will require updating the sorted map which might take O(log n) time.

Time: inc/dec: O(log n), getMaxKey/getMinKey: O(1)Space: O(n)

Trade-offs: This approach is simpler to implement but does not meet the O(1) average time complexity requirement for inc/dec. It is preferable if inc/dec operations are less frequent than getMaxKey/getMinKey operations and a slight increase in time complexity is acceptable.

✨

Key Insights

1Insight 1: Maintaining a doubly-linked list of buckets, where each bucket stores keys with the same count, allows for O(1) access to minimum and maximum counts.
2Insight 2: A hash map mapping keys to their respective nodes in the linked list enables O(1) increment and decrement operations.
3Insight 3: Correctly handling edge cases such as an empty AllOne data structure or the first or last node being updated is crucial for correct functionality.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Linked List

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

Edge Cases

Edge case 1: Empty `AllOne` structure: `getMaxKey` and `getMinKey` must return an empty string.
Edge case 2: Decrementing a key to zero: The key must be removed from the `key_to_node` map and its current node.
Edge case 3: Incrementing a new key: A new node with count 1 must be created and inserted after the head.
Edge case 4: Incrementing and Decrementing the only key in a bucket: Correctly handles removal of empty buckets.
Edge case 5: Incrementing/decrementing adjacent buckets: Correctly merges keys from adjacent buckets if needed to maintain buckets of same count.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the `key_to_node` map when incrementing or decrementing a key.

✗Common mistake 2: Failing to remove a key from the `key_to_node` map when its count becomes zero.

✗Common mistake 3: Incorrectly handling edge cases such as an empty `AllOne` structure, or the first/last node being updated.

✗Common mistake 4: Not freeing the memory in C/C++ properly which would lead to memory leaks.

✗Common mistake 5: Not considering the case where a node becomes empty after incrementing or decrementing a key. These nodes must be removed from the doubly linked list.

✅

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: Tracking the popularity of search queries, where frequent queries have higher counts.

🏦Real-world scenario 2: Analyzing website traffic to identify the most and least visited pages.

🎮Real-world scenario 3: Maintaining a leader board in a game, tracking player scores and identifying players with the highest and lowest scores.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: August 12, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "All O`one Data Structure"?

For this hard-level Hash Table 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.

🤖 Need help with this problem?