What is the time complexity of Encode and Decode TinyURL?

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

What data structures are used to solve Encode and Decode TinyURL?

This problem uses Hash Table, String, Design, Hash Function. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Encode and Decode TinyURL?

Insight 1: We need a way to store the mapping between long URLs and short URLs. Hash tables (dictionaries in Python, HashMaps in Java/C++) are ideal for this. Insight 2: The short URLs should be unique. A simple counter can be used to generate unique identifiers. These identifiers can be converted to a base-N representation (using alphanumeric characters) to create shorter, more user-friendly URLs.

Which companies ask LeetCode 535. Encode and Decode TinyURL in interviews?

This problem is frequently asked at Shopify.

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Encode and Decode TinyURL - LeetCode 535 Solution

Encode and Decode TinyURL - Complete Solution Guide

Encode and Decode TinyURL is LeetCode problem 535, 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

Note: This is a companion problem to the System Design problem: Design TinyURL . TinyURL is a URL shortening service where you enter a URL such as https://leetcode.com/problems/design-tinyurl and it returns a short URL such as http://tinyurl.com/4e9iAk . Design a class to encode a URL and decode a tiny URL. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL. I

Detailed Explanation

The problem asks us to design a system similar to TinyURL, a URL shortening service. We need to implement two functions: `encode(longUrl)` which takes a long URL as input and returns a shortened URL, and `decode(shortUrl)` which takes a shortened URL and returns the original long URL. There are no specific constraints on the encoding/decoding algorithm, but the solution must guarantee that the decoded URL is the same as the original URL.

Solution Approach

The provided solutions use a combination of a hash table and a counter. The `encode` function checks if the long URL already exists in the hash table. If it does, it returns the corresponding short URL. If not, it increments a counter, converts the counter value to a base-62 representation (using alphanumeric characters), stores the mapping in both `long_to_short` and `short_to_long` hash tables, and returns the new short URL. The `decode` function simply looks up the short URL in the `short_to_long` hash table and returns the corresponding long URL.

Step-by-Step Algorithm

Step 1: Initialize two hash tables (dictionaries): `long_to_short` to store the mapping from long URLs to short keys and `short_to_long` to store the mapping from short keys to long URLs. Also, initialize a counter to generate unique short keys and define the set of characters (`alphabet`) to be used for base-N conversion.
Step 2: In the `encode` function, check if the long URL already exists in the `long_to_short` hash table. If it does, return the corresponding short URL by concatenating the base URL with short key.
Step 3: If the long URL is not in the hash table, increment the counter.
Step 4: Convert the counter value to a base-N representation using the `alphabet`. This generates a unique short key.
Step 5: Store the mapping between the long URL and the short key in both hash tables.
Step 6: Return the newly generated short URL by concatenating the base URL with the short key.
Step 7: In the `decode` function, extract the short key from the short URL by removing the base URL prefix.
Step 8: Look up the short key in the `short_to_long` hash table and return the corresponding long URL.

Key Insights

Insight 1: We need a way to store the mapping between long URLs and short URLs. Hash tables (dictionaries in Python, HashMaps in Java/C++) are ideal for this.
Insight 2: The short URLs should be unique. A simple counter can be used to generate unique identifiers. These identifiers can be converted to a base-N representation (using alphanumeric characters) to create shorter, more user-friendly URLs.
Insight 3: The length of the short URL is related to the base (number of characters used) and the number of URLs we expect to store. A larger base allows for shorter URLs for the same number of URLs stored.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Design, Hash Function.

Companies

Asked at: Shopify.

Encode and Decode TinyURL | LeetCode 535

Medium

Hash Table String Design Hash Function

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

public class Codec {

    private Map<String, String> longToShort = new HashMap<>();
    private Map<String, String> shortToLong = new HashMap<>();
    private String alphabet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
    private String baseUrl = "http://tinyurl.com/";
    private int counter = 0;

    // Encodes a URL to a shortened URL.
    public String encode(String longUrl) {
        if (longToShort.containsKey(longUrl)) {
            return baseUrl + longToShort.get(longUrl);
        }

        counter++;
        int num = counter;

        StringBuilder keyBuilder = new StringBuilder();
        int base = alphabet.length();

        while (num > 0) {
            int rem = num % base;
            num = num / base;
            keyBuilder.append(alphabet.charAt(rem));
        }

        String shortKey = keyBuilder.reverse().toString();

        longToShort.put(longUrl, shortKey);
        shortToLong.put(shortKey, longUrl);

        return baseUrl + shortKey;
    }

    // Decodes a shortened URL to its original URL.
    public String decode(String shortUrl) {
        String shortKey = shortUrl.replace(baseUrl, "");
        return shortToLong.get(shortKey);
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(url));

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Problem Info

Description

Problem 535: Encode and Decode TinyURL - Detailed explanation and solution approaches.

Topics

Hash Table String Design Hash Function

Companies

Shopify

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:535

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 hash tables (dictionaries): `long_to_short` to store the mapping from long URLs to short keys and `short_to_long` to store the mapping from short keys to long URLs. Also, initialize a counter to generate unique short keys and define the set of characters (`alphabet`) to be used for base-N conversion.
2
Step 2: In the `encode` function, check if the long URL already exists in the `long_to_short` hash table. If it does, return the corresponding short URL by concatenating the base URL with short key.
3
Step 3: If the long URL is not in the hash table, increment the counter.
4
Step 4: Convert the counter value to a base-N representation using the `alphabet`. This generates a unique short key.
5
Step 5: Store the mapping between the long URL and the short key in both hash tables.
6
Step 6: Return the newly generated short URL by concatenating the base URL with the short key.
7
Step 7: In the `decode` function, extract the short key from the short URL by removing the base URL prefix.
8
Step 8: Look up the short key in the `short_to_long` hash table and return the corresponding long URL.

💡

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 dominated by hash table operations, which are O(1) on average. The conversion from the counter to a base-N representation takes logarithmic time relative to the counter's current value, but this is still effectively constant in the context of each individual 'encode' operation. The space complexity is linear because we store each unique URL in the hash tables.

📖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 Random String Generator

Instead of using a counter, a random string of a fixed length can be generated as the short key. This avoids the need for a counter but introduces the possibility of collisions (generating the same short key for different long URLs).

Time: O(1) for encode and decode on average, assuming collision resolution is efficient.Space: O(n)

Trade-offs: This approach can lead to collisions, requiring a retry mechanism to generate a unique short URL. The collision probability depends on the length of the random string and the number of URLs stored. This approach might be faster than counter approach if collisions are rare.

Database with Auto-Increment ID

Using a database with an auto-incrementing primary key. The long URL can be stored in the database, and the auto-incremented ID can be used as the short key. This guarantees uniqueness and persistence of the URL mappings.

Time: O(1) for encode and decode, assuming database operations are constant time.Space: O(n), where n is the number of URLs stored in the database.

Trade-offs: This approach requires a database setup, which adds complexity. However, it provides persistence and scalability for large-scale URL shortening services.

✨

Key Insights

1Insight 1: We need a way to store the mapping between long URLs and short URLs. Hash tables (dictionaries in Python, HashMaps in Java/C++) are ideal for this.
2Insight 2: The short URLs should be unique. A simple counter can be used to generate unique identifiers. These identifiers can be converted to a base-N representation (using alphanumeric characters) to create shorter, more user-friendly URLs.
3Insight 3: The length of the short URL is related to the base (number of characters used) and the number of URLs we expect to store. A larger base allows for shorter URLs for the same number of URLs stored.

📚

Key Terms Explained

For Students

Hash Table

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

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Very long URLs: The solution works correctly for long URLs as the hash tables can store strings of any length (within reasonable memory limits). The code uses `strlen` in C, so extremely large urls can lead to integer overflows and should be handled appropriately.
Edge case 2: Duplicate URLs: The solution efficiently handles duplicate URLs by checking if the URL already exists in the `long_to_short` hash table before generating a new short URL. This prevents unnecessary generation of new short URLs for the same long URL.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling collisions when using a random string generator. It's crucial to check if the generated short key already exists and regenerate it if necessary.

✗Common mistake 2: Not considering the size of the short key. If the short key is too short, collisions will become more frequent as the number of URLs increases. A longer short key requires more storage space but reduces the probability of collisions.

✗Common mistake 3: Not handling the base URL properly. The code needs to correctly add and remove the base URL when encoding and decoding.

✅

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: URL shortening services like TinyURL and Bitly use similar algorithms to generate short URLs for sharing on social media or in emails.

🏦Real-world scenario 2: Systems that generate unique identifiers for database records or other objects can use a similar base-N encoding scheme to create shorter, more human-readable identifiers.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 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: November 23, 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 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Encode and Decode TinyURL"?

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

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