Encode and Decode TinyURL

Medium

Amazon

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Topics:

StringsArraysDatabase Problems

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.

Specifically:

Implement a Solution class with the following methods:
- Solution(): Initializes the object of the system.
- String encode(String longUrl): Returns a tiny URL for the given longUrl.
- String decode(String shortUrl): Returns the original long URL for the given shortUrl. It is guaranteed that the given shortUrl was encoded by the same object.
Provide example usage to demonstrate encoding and decoding.
Explain the time and space complexity of your approach.
Discuss potential edge cases and how to handle them.

For instance, if the input is "https://leetcode.com/problems/design-tinyurl", the output should be the same URL after encoding and decoding. Provide code examples in Python.

Encode and Decode TinyURL

Medium

Amazon

2 views

Topics:

StringsArraysDatabase Problems

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.

Specifically:

Implement a Solution class with the following methods:
- Solution(): Initializes the object of the system.
- String encode(String longUrl): Returns a tiny URL for the given longUrl.
- String decode(String shortUrl): Returns the original long URL for the given shortUrl. It is guaranteed that the given shortUrl was encoded by the same object.
Provide example usage to demonstrate encoding and decoding.
Explain the time and space complexity of your approach.
Discuss potential edge cases and how to handle them.

For instance, if the input is "https://leetcode.com/problems/design-tinyurl", the output should be the same URL after encoding and decoding. Provide code examples in Python.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected lifespan or persistence of the encoded TinyURLs? Should they be stored indefinitely, or is there a time limit?
How many unique long URLs can we expect to encode over time? This helps determine the necessary length and complexity of the encoding scheme.
Are we concerned about collisions, where different long URLs map to the same TinyURL? If so, how should collisions be handled?
What characters are allowed in the TinyURL? Are there any restrictions on the character set?
What should happen if the same long URL is encoded multiple times? Should it always return the same TinyURL, or can a new one be generated?

Brute Force Solution

Approach

The goal is to make a long web address shorter and then be able to get back the original web address from the short one. The brute force approach involves trying simple short codes until one that isn't already used is found.

Here's how the algorithm would work step-by-step:

When someone gives you a long web address to shorten, start by trying a very simple short code, like just the letter 'a'.
Check if that short code ('a') is already in use for another web address.
If 'a' is already taken, try 'b'. Keep going through the alphabet ('c', 'd', 'e', etc.).
If you run out of single letters, try combinations of two letters, like 'aa', 'ab', 'ac', and so on.
Keep increasing the length and complexity of the short code until you find one that hasn't been used before.
Once you find an unused short code, save the connection between that short code and the original long web address in a safe place.
When someone gives you a short code, look up the original long web address that is saved with it.

Code Implementation

class TinyURL:

    def __init__(self):
        self.url_to_short_url = {}
        self.short_url_to_url = {}
        self.characters = "abcdefghijklmnopqrstuvwxyz"
        self.short_url_length = 1

    def encode(self, long_url):
        short_url = self.generate_short_url()

        # Ensure that the generated short_url is unique
        while short_url in self.short_url_to_url:
            short_url = self.generate_short_url()

        self.url_to_short_url[long_url] = short_url
        self.short_url_to_url[short_url] = long_url
        return short_url

    def decode(self, short_url):
        return self.short_url_to_url.get(short_url)

    def generate_short_url(self):
        short_url = self.generate_candidate(self.short_url_length)

        if self.is_exhausted(self.short_url_length):
            self.short_url_length += 1

        return short_url

    def generate_candidate(self, current_length):
        if current_length == 1:
            for char in self.characters:
                if char not in self.short_url_to_url:
                    return char
            return 'aa'
        else:
            #This loop creates short urls of a given length, 'aa', 'ab', etc
            generated_short_urls = self.generate_all_strings(current_length)

            for short_url_candidate in generated_short_urls:
                if short_url_candidate not in self.short_url_to_url:
                    return short_url_candidate

            return self.generate_all_strings(current_length + 1)[0]

    def generate_all_strings(self, string_length):
        if string_length == 0:
            return ['']

        previous_strings = self.generate_all_strings(string_length - 1)
        all_strings = []

        #Iterate and append
        for previous_string in previous_strings:
            for character in self.characters:
                all_strings.append(previous_string + character)

        return all_strings

    def is_exhausted(self, current_length):
        number_of_possible_urls = len(self.characters) ** current_length

        # This ensures that we do not create duplicate short urls.
        if number_of_possible_urls <= len(self.short_url_to_url):
            return True
        return False

Big(O) Analysis

Time Complexity

O(n) – Encoding involves generating short codes of increasing length (a, b, c, ... aa, ab, ...), and for each generated code, checking if it already exists. In the worst case, we might need to generate and check 'n' short codes where 'n' is the number of previously encoded URLs. Each check of whether a short code exists takes constant time, O(1), using a hash map. Decoding a short URL involves a single lookup in the hash map, which also takes O(1) time. Because encoding in the worst case can check 'n' URLs, the total operations are proportional to 'n'. Thus, the overall complexity is O(n).

Space Complexity

O(N) – The primary space complexity arises from storing the mapping between long URLs and their corresponding short URLs. In the worst-case scenario, we might need to store N long URLs and their generated short URLs where N represents the number of long URLs encoded. This storage, likely implemented using a hash map or similar data structure, necessitates memory proportional to the number of encoded URLs. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The key is to create a system that can convert long URLs into short, unique identifiers and back again. We'll use a combination of a storage system and a way to create those unique identifiers.

Here's how the algorithm would work step-by-step:

First, we need a way to remember which short identifier goes with which long URL, almost like a dictionary.
When we get a long URL that needs shortening, we'll check if we already have a short identifier for it. If we do, we just return that.
If it's a brand new long URL, we'll create a unique, short identifier for it. We can do this by incrementing a counter and converting it into a short string using letters and numbers.
Then, we store the long URL along with its new short identifier in our dictionary.
When someone gives us a short identifier and wants the original long URL, we simply look it up in our dictionary and return it.
If we cannot find the short identifier in our dictionary, it's probably an invalid short URL, and we can return nothing.

Code Implementation

class TinyURL:
    def __init__(self):
        self.url_to_short_url_map = {}
        self.short_url_to_url_map = {}
        self.counter = 0
        self.base_url = "http://tinyurl.com/"

    def encode(self, long_url):
        # Check if long_url is already encoded
        if long_url in self.url_to_short_url_map:
            return self.url_to_short_url_map[long_url]

        # Generate a unique short URL
        self.counter += 1
        short_url_id = self.convert_to_base62(self.counter)
        short_url = self.base_url + short_url_id

        # Store mapping
        self.url_to_short_url_map[long_url] = short_url
        self.short_url_to_url_map[short_url] = long_url

        return short_url

    def decode(self, short_url):
        # Retrieve the original URL using the short URL.
        if short_url in self.short_url_to_url_map:
            return self.short_url_to_url_map[short_url]
        else:
            return None

    def convert_to_base62(self, number):
        # Use a base62 conversion for creating unique identifiers
        characters = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
        base = len(characters)
        result = []

        while number > 0:
            remainder = number % base
            result.append(characters[remainder])
            number //= base

        return ''.join(reversed(result))

Big(O) Analysis

Time Complexity

O(1) – The encode operation involves checking if a long URL exists in a dictionary (average case O(1)), generating a short key which is an arithmetic operation (O(1)), and storing the key-value pair in the dictionary (average case O(1)). The decode operation involves looking up a short URL in the dictionary (average case O(1)). Therefore, both encode and decode operations have constant time complexity.

Space Complexity

O(N) – The primary space usage comes from the dictionary that stores the mapping between short URLs and long URLs. In the worst case, we might encode N distinct long URLs, each requiring a new entry in the dictionary. Therefore, the space required to store this mapping grows linearly with the number of encoded URLs, N. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty long URL provided as input	Return null or throw an exception indicating invalid input, as there is nothing to encode.
Very long URL (exceeding typical URL length limits)	Ensure the database or storage used for mapping can handle extremely long URLs, and consider truncating or hashing if necessary.
URL already exists in the database	Either return the existing tiny URL or generate a new tiny URL to avoid conflicts depending on requirements.
Running out of available tiny URLs	Implement a mechanism to expand the tiny URL character set or length if the available namespace is exhausted.
Collision of generated tiny URLs	Use a robust collision resolution strategy, such as retrying with a different random seed or expanding the tiny URL length.
Decoding a non-existent tiny URL	Return null or an error message indicating that the tiny URL is not found in the database.
Malicious long URL input (e.g., containing script tags or other exploits)	Implement input validation and sanitization to prevent potential security vulnerabilities.
High volume of requests (scalability)	Ensure the encoding and decoding service can handle a large number of requests concurrently by using appropriate data structures and potentially sharding the database.