Problem Description

This problem asks you to design a URL shortening service similar to TinyURL. You need to create a system that can convert long URLs into short URLs and then convert those short URLs back to their original long URLs.

The task requires implementing a Codec class with the following functionality:

1. Constructor (__init__): Initializes the URL shortening system
2. encode(longUrl): Takes a long URL as input (like https://leetcode.com/problems/design-tinyurl) and returns a shortened version (like http://tinyurl.com/4e9iAk)
3. decode(shortUrl): Takes a shortened URL and returns the original long URL

The key requirements are:

• Every long URL should be encoded to a unique short URL
• Every short URL should decode back to its original long URL
• The same object instance will be used for both encoding and decoding operations
• There are no restrictions on the specific algorithm you use for encoding/decoding

The solution uses a simple counter-based approach:

• Maintains a dictionary (m) to store mappings between IDs and long URLs
• Uses an incrementing counter (idx) to generate unique IDs for each new URL
• Creates short URLs by appending the ID to a base domain (https://tinyurl.com/)
• When decoding, extracts the ID from the short URL and looks up the original URL in the dictionary

For example:

• Input: encode("https://leetcode.com/problems/design-tinyurl")
• Output: "https://tinyurl.com/1" (for the first URL)
• Input: decode("https://tinyurl.com/1")
• Output: "https://leetcode.com/problems/design-tinyurl"

Intuition

The core challenge in URL shortening is creating a bidirectional mapping between long URLs and short identifiers. We need a way to generate unique short codes and remember which long URL each code corresponds to.

The simplest approach is to think of this as a database problem. When we receive a long URL, we assign it a unique ID and store the mapping. This leads us to consider using a counter that increments with each new URL - it's guaranteed to be unique and simple to implement.

Why use a counter instead of other approaches like hashing? Hashing might seem natural, but it has complications:

• Hash collisions would require handling
• We'd need to store the hash-to-URL mapping anyway
• The hash output might be longer than necessary

A sequential counter gives us the shortest possible unique identifiers: 1, 2, 3, etc. This minimizes the length of our short URLs.

For storage, we need a data structure that allows fast lookups when decoding. A dictionary (hash map) is perfect - it gives us O(1) lookup time. We use the counter value as the key and the long URL as the value.

The encoding process becomes:

1. Increment the counter to get a new unique ID
2. Store the ID-to-URL mapping in our dictionary
3. Return the short URL by appending the ID to our base domain

The decoding process is even simpler:

1. Extract the ID from the short URL (the part after the last /)
2. Look up the ID in our dictionary to get the original URL

This approach trades simplicity for features. While more sophisticated solutions might handle custom aliases, analytics, or URL deduplication, this counter-based method solves the core problem with minimal complexity and guaranteed correctness.

Solution Approach

The implementation uses a counter-based approach with a hash map for storage. Let's walk through each component:

Data Structure Setup:

• self.m = defaultdict(): A dictionary to store the mapping between IDs and long URLs
• self.idx = 0: A counter that starts at 0 and increments for each new URL
• self.domain = 'https://tinyurl.com/': The base domain for all shortened URLs

Encoding Algorithm:

def encode(self, longUrl: str) -> str:
    self.idx += 1
    self.m[str(self.idx)] = longUrl
    return f'{self.domain}{self.idx}'

The encoding process:

1. Increment the counter self.idx by 1 to generate a new unique ID
2. Store the mapping in the dictionary with the ID (converted to string) as the key and the long URL as the value
3. Construct the short URL by concatenating the domain with the ID using an f-string
4. Return the complete short URL (e.g., 'https://tinyurl.com/1')

Decoding Algorithm:

def decode(self, shortUrl: str) -> str:
    idx = shortUrl.split('/')[-1]
    return self.m[idx]

The decoding process:

1. Extract the ID from the short URL by splitting on / and taking the last element
   • For example: 'https://tinyurl.com/1'.split('/') gives ['https:', '', 'tinyurl.com', '1']
   • Taking [-1] gives us '1'
2. Use the extracted ID to look up the original URL in the dictionary
3. Return the original long URL

Time and Space Complexity:

• encode(): O(1) time - incrementing counter and dictionary insertion are constant time operations
• decode(): O(1) time - string splitting is O(k) where k is the number of / characters (constant for our URLs), and dictionary lookup is O(1)
• Space: O(n) where n is the number of unique URLs encoded, as we store each mapping in the dictionary

This solution guarantees no collisions since each URL gets a unique, monotonically increasing ID. The trade-off is that the short URLs are predictable and sequential, which might not be desirable in a real-world system where security or unpredictability is important.

Example Walkthrough

Let's walk through encoding and decoding two URLs to see how the counter-based approach works:

Initial State:

• idx = 0
• m = {} (empty dictionary)
• domain = 'https://tinyurl.com/'

Step 1: Encode first URL

• Input: encode("https://leetcode.com/problems/design-tinyurl")
• Process:
  1. Increment idx: 0 → 1
  2. Store in dictionary: m['1'] = "https://leetcode.com/problems/design-tinyurl"
  3. Create short URL: 'https://tinyurl.com/' + '1'
• Output: "https://tinyurl.com/1"
• State after: idx = 1, m = {'1': 'https://leetcode.com/problems/design-tinyurl'}

Step 2: Encode second URL

• Input: encode("https://example.com/very/long/path")
• Process:
  1. Increment idx: 1 → 2
  2. Store in dictionary: m['2'] = "https://example.com/very/long/path"
  3. Create short URL: 'https://tinyurl.com/' + '2'
• Output: "https://tinyurl.com/2"
• State after: idx = 2, m = {'1': 'https://leetcode.com/problems/design-tinyurl', '2': 'https://example.com/very/long/path'}

Step 3: Decode first short URL

• Input: decode("https://tinyurl.com/1")
• Process:
  1. Split URL: "https://tinyurl.com/1".split('/') → ['https:', '', 'tinyurl.com', '1']
  2. Extract ID: Take last element → '1'
  3. Lookup in dictionary: m['1'] → "https://leetcode.com/problems/design-tinyurl"
• Output: "https://leetcode.com/problems/design-tinyurl"

Step 4: Decode second short URL

• Input: decode("https://tinyurl.com/2")
• Process:
  1. Split URL and get last element → '2'
  2. Lookup: m['2'] → "https://example.com/very/long/path"
• Output: "https://example.com/very/long/path"

The counter ensures each URL gets a unique ID (1, 2, 3...), and the dictionary provides instant lookup for decoding.

Solution Implementation

Time and Space Complexity

Time Complexity:

• encode(): O(1) - The method performs constant time operations including incrementing a counter, storing a key-value pair in a dictionary, and string formatting.
• decode(): O(1) - The method performs a string split operation which takes O(n) where n is the length of the URL, but since the URL length is bounded and relatively small, we can consider it O(1). Dictionary lookup is also O(1) on average.

Space Complexity:

• Overall: O(n) where n is the number of unique URLs encoded.
• The dictionary self.m stores a mapping for each unique long URL that has been encoded, growing linearly with the number of URLs.
• Each encode() call adds one entry to the dictionary, consuming O(1) additional space per call.
• The decode() method doesn't use any additional space beyond temporary variables, so it's O(1) for the operation itself.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. URL Collision in High-Traffic Scenarios

The counter-based approach is not thread-safe. In a multi-threaded environment or distributed system, multiple requests could read the same counter value before incrementing, causing collisions where different long URLs get mapped to the same short URL.

Solution:

import threading

class Codec:
    def __init__(self):
        self.url_map = defaultdict(str)
        self.counter = 0
        self.base_domain = 'https://tinyurl.com/'
        # Add a lock for thread safety
        self.lock = threading.Lock()

    def encode(self, longUrl: str) -> str:
        with self.lock:
            self.counter += 1
            current_id = self.counter
        self.url_map[str(current_id)] = longUrl
        return f'{self.base_domain}{current_id}'

2. Predictable URL Pattern Security Risk

Sequential IDs (1, 2, 3...) make it trivial for malicious users to enumerate all shortened URLs by simply incrementing the ID, potentially exposing private or sensitive URLs.

Solution:

import random
import string

class Codec:
    def __init__(self):
        self.url_to_code = {}
        self.code_to_url = {}
        self.base_domain = 'https://tinyurl.com/'
      
    def encode(self, longUrl: str) -> str:
        # Check if URL was already encoded
        if longUrl in self.url_to_code:
            return f'{self.base_domain}{self.url_to_code[longUrl]}'
      
        # Generate random 6-character code
        while True:
            code = ''.join(random.choices(string.ascii_letters + string.digits, k=6))
            if code not in self.code_to_url:
                break
      
        self.url_to_code[longUrl] = code
        self.code_to_url[code] = longUrl
        return f'{self.base_domain}{code}'

3. Missing Duplicate URL Handling

The current implementation creates a new short URL every time the same long URL is encoded, wasting storage and creating multiple short URLs for the same resource.

Solution:

class Codec:
    def __init__(self):
        self.url_map = defaultdict(str)
        self.reverse_map = {}  # Map from long URL to short URL ID
        self.counter = 0
        self.base_domain = 'https://tinyurl.com/'

    def encode(self, longUrl: str) -> str:
        # Check if this URL was already encoded
        if longUrl in self.reverse_map:
            return f'{self.base_domain}{self.reverse_map[longUrl]}'
      
        self.counter += 1
        self.url_map[str(self.counter)] = longUrl
        self.reverse_map[longUrl] = str(self.counter)
        return f'{self.base_domain}{self.counter}'

4. No Input Validation

The code doesn't validate if the input URLs are properly formatted or if the short URL in decode() actually exists in the system.

Solution:

class Codec:
    def decode(self, shortUrl: str) -> str:
        # Validate short URL format
        if not shortUrl.startswith(self.base_domain):
            raise ValueError(f"Invalid short URL format: {shortUrl}")
      
        url_id = shortUrl.split('/')[-1]
      
        # Check if ID exists in our system
        if url_id not in self.url_map:
            raise KeyError(f"Short URL not found: {shortUrl}")
      
        return self.url_map[url_id]

5. Memory Limitations for Large Scale

Using a simple in-memory dictionary means all mappings are lost if the service restarts, and memory usage grows unbounded with the number of URLs.

Solution: Consider using a persistent database (Redis, PostgreSQL) or implementing an LRU cache with database backing:

from functools import lru_cache

class Codec:
    def __init__(self, max_cache_size=10000):
        self.cache = {}  # In production, use Redis or a database
        self.counter = self._load_counter()  # Load from persistent storage
      
    @lru_cache(maxsize=1000)
    def decode(self, shortUrl: str) -> str:
        url_id = shortUrl.split('/')[-1]
        # In production, fetch from database if not in cache
        return self._fetch_from_storage(url_id)