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Encode and Decode TinyURL #3 Most Asked

Medium

Shopify

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Topics:

ArraysStringsHash TableDesign

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.

Implement the Solution class:

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.

Example 1:

Input: url = "https://leetcode.com/problems/design-tinyurl"
Output: "https://leetcode.com/problems/design-tinyurl"

Explanation:
Solution obj = new Solution();
string tiny = obj.encode(url); // returns the encoded tiny url.
string ans = obj.decode(tiny); // returns the original url after decoding it.

Constraints:

1 <= url.length <= 10⁴
url is guranteed to be a valid URL.

Try coding it in LeetCode

Encode and Decode TinyURL #3 Most Asked

Medium

Shopify

8 views

Topics:

ArraysStringsHash TableDesign

Note: This is a companion problem to the System Design problem: Design TinyURL.

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.

Implement the Solution class:

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.

Example 1:

Input: url = "https://leetcode.com/problems/design-tinyurl"
Output: "https://leetcode.com/problems/design-tinyurl"

Explanation:
Solution obj = new Solution();
string tiny = obj.encode(url); // returns the encoded tiny url.
string ans = obj.decode(tiny); // returns the original url after decoding it.

Constraints:

1 <= url.length <= 10⁴
url is guranteed to be a valid URL.

Try coding it in LeetCode

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:

If the `encode` method is called multiple times with the same long URL, should it consistently return the same short URL, or generate a new one each time?
What are the specific format requirements for the short URL? For instance, should it use the prefix `http://tinyurl.com/`, and are there any constraints on the length or character set (e.g., alphanumeric) for the unique code?
My encoding strategy must generate a unique code for each URL. If I use an approach like hashing or random generation, collisions are possible. How should my algorithm handle such collisions?
Is the mapping between long and short URLs expected to be persistent, or is it sufficient for the mapping to exist only for the lifetime of a single `Solution` object instance?
The problem guarantees that a `shortUrl` given to `decode` was previously generated. Does this also mean I can assume the format of the `shortUrl` string itself is always valid and doesn't require parsing validation?

Brute Force Solution

Approach

The simplest approach is to generate a random short code for each long web address. If we happen to generate a code that we've already used, we just keep trying to generate new random codes until we find one that's available. We'll maintain a record of which short code corresponds to which long address.

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

To shorten a long web address, start by creating a short, random sequence of characters.
Next, look through all the short links you've already created to see if this new random one has been used before.
If you find that it's already in use, you must throw it away and create another random sequence from scratch.
Keep repeating this process of creating a random sequence and checking if it's taken, over and over, until you finally make one that is completely new.
Once you have a unique new short link, record it alongside the original long web address so you can remember which is which.
When you need to find the original address from a short link, you just look up the short link in your records to find the long address you saved with it.

Code Implementation

class Codec:

    def __init__(self):
        self.url_database = []
        self.alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
        self.short_url_length = 6

    def encode(self, longUrl: str) -> str:
        while True:
            # Keep generating random names until we find one that is not already in our database.

            potential_short_name = "".join([random.choice(self.alphabet) for _ in range(self.short_url_length)])
            
            is_name_taken = False
            for _, existing_short_name in self.url_database:
                if existing_short_name == potential_short_name:
                    is_name_taken = True
                    break

            if not is_name_taken:
                # Once an unused short name is found, we record the pair in our 'notebook' for future lookups.

                self.url_database.append((longUrl, potential_short_name))
                
                return "http://tinyurl.com/" + potential_short_name

    def decode(self, shortUrl: str) -> str:
        short_code_to_find = shortUrl.replace("http://tinyurl.com/", "")
        
        # To find the original address, we must look through every entry one-by-one, just like a notebook.

        for original_address, short_code in self.url_database:
            
            if short_code == short_code_to_find:
                return original_address
        
        return ""

Big(O) Analysis

Time Complexity

O(n) – Let n be the number of URLs that have already been encoded. The runtime is driven by the encode operation's collision handling, where it must generate a new random code if the current one is already taken. In the worst-case scenario, the system could unluckily generate codes that match all n existing entries before finally creating a unique one. This repetition means the process of generating and checking a code may need to be performed a number of times proportional to n. This results in a worst-case time complexity of O(n), while the decode operation remains a constant O(1) lookup.

Space Complexity

O(N * L) – The primary auxiliary space is the 'record' used to store the mapping between short links and their corresponding long web addresses. This data structure, typically a hash map, must hold an entry for every unique URL that is encoded. If N is the number of URLs stored and L is the average length of a URL, the space grows linearly with the total size of all stored URLs. Therefore, the space complexity is O(N * L).

Optimal Solution

Approach

The strategy is to assign a unique, ever-increasing number to each new long web address, much like a ticket at a coat check. This number is then converted into a very short string, and we simply remember which short string goes with which long address, guaranteeing no two addresses get the same short version.

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

Think of the system as having a simple counter, like a ticket dispenser, that starts at one and goes up.
When a new long web address comes in, we give it the next available number from our counter.
We then convert this number into a very compact code using a special character set that includes both letters and digits. This keeps the resulting code extremely short.
We store this connection: the original long address is now linked to this new, unique code.
Finally, we increment our counter, so it's ready with a fresh number for the next web address.
To get the long address back from a short code, we simply do the reverse. We translate the code back into its original number.
With that number, we can immediately look up and find the original long web address that we saved.

Code Implementation

class Codec:

    def __init__(self):
        self.alphabet_characters = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
        self.base_url_prefix = "http://tinyurl.com/"
        self.long_urls_list = []
        self.url_to_short_code_map = {}

    def encode(self, long_url: str) -> str:
        # To avoid creating duplicates, we first check if this URL already has a short version.

        if long_url in self.url_to_short_code_map:
            return self.url_to_short_code_map[long_url]

        # Storing the new URL gives it a unique, sequential ID based on its list position.

        self.long_urls_list.append(long_url)
        unique_id_number = len(self.long_urls_list) - 1

        # The ID is converted to a compact base-62 representation for the short URL code.

        short_identifier_code = ""
        if unique_id_number == 0:
            short_identifier_code = self.alphabet_characters[0]
        else:
            temporary_id_number = unique_id_number
            while temporary_id_number > 0:
                remainder_index = temporary_id_number % 62
                short_identifier_code = self.alphabet_characters[remainder_index] + short_identifier_code
                temporary_id_number //= 62
        
        full_short_url = self.base_url_prefix + short_identifier_code
        self.url_to_short_code_map[long_url] = full_short_url
        return full_short_url

    def decode(self, short_url: str) -> str:
        short_identifier_code = short_url.replace(self.base_url_prefix, "")
        unique_id_number = 0

        # The short code must be converted back to its original numeric ID to find the long URL.

        for character_in_code in short_identifier_code:
            unique_id_number = unique_id_number * 62 + self.alphabet_characters.find(character_in_code)

        # The decoded ID directly corresponds to the index of the original URL in our master list.

        return self.long_urls_list[unique_id_number]

Big(O) Analysis

Time Complexity

O(n) – The time complexity for both encode and decode is determined by the length of the input string, let's call it n. For the encode function, the dominant operation is storing the long URL, which takes time proportional to its length, n. For the decode function, the cost is driven by iterating through the short URL's characters to convert it back to its original number, which is also an O(n) operation. Since all other steps, like map lookups and counter increments, are constant time on average, the overall complexity is O(n).

Space Complexity

O(N) – The primary driver of space complexity is the data structure used to 'remember which short string goes with which long address'. This storage, typically a hash map or a list, holds a mapping for every unique long URL that has been encoded. If N represents the number of unique URLs processed by the system, this storage will grow to hold N entries. Therefore, the auxiliary space required is directly proportional to N.

Edge Cases

Case	How to Handle
The encode method is called multiple times with the exact same longUrl.	The implementation should return the pre-existing shortUrl for that longUrl rather than creating a new one to ensure idempotency.
The decode method receives a full URL string, including the base path like 'http://tinyurl.com/', not just the unique code.	The solution must parse the input string to extract only the unique short code before performing the lookup.
The number of unique URLs to encode exhausts all possible combinations for a fixed-length short code.	A counter-based approach handles this by naturally increasing the length of the generated code as the counter value grows.
Two distinct longUrls map to the same shortUrl code, causing a collision.	A collision must be detected and resolved, for example by regenerating the code until a unique one is found.
The input longUrl is at the maximum allowed length of 10,000 characters.	The data structure, typically a hash map, must efficiently store and retrieve these large strings without significant performance degradation.
The encode method is called with a URL that is itself a previously generated short URL.	This input should be treated like any other valid URL and encoded to a new, different short URL.
An internal counter used for generating unique IDs exceeds the maximum value for its data type, like Integer.MAX_VALUE.	Using a 64-bit integer type (long) for the counter provides a much larger keyspace to prevent overflow for practical purposes.
The data structures for forward (long-to-short) and reverse (short-to-long) lookups become inconsistent.	Both mappings must be updated atomically within the encode function to ensure that every encoded URL is decodable.

The encode method is called multiple times with the exact same longUrl.

How to Handle:

The implementation should return the pre-existing shortUrl for that longUrl rather than creating a new one to ensure idempotency.

The decode method receives a full URL string, including the base path like 'http://tinyurl.com/', not just the unique code.

How to Handle:

The solution must parse the input string to extract only the unique short code before performing the lookup.

The number of unique URLs to encode exhausts all possible combinations for a fixed-length short code.

How to Handle:

A counter-based approach handles this by naturally increasing the length of the generated code as the counter value grows.

Two distinct longUrls map to the same shortUrl code, causing a collision.

How to Handle:

A collision must be detected and resolved, for example by regenerating the code until a unique one is found.

The input longUrl is at the maximum allowed length of 10,000 characters.

How to Handle:

The data structure, typically a hash map, must efficiently store and retrieve these large strings without significant performance degradation.

The encode method is called with a URL that is itself a previously generated short URL.

How to Handle:

This input should be treated like any other valid URL and encoded to a new, different short URL.

An internal counter used for generating unique IDs exceeds the maximum value for its data type, like Integer.MAX_VALUE.

How to Handle:

Using a 64-bit integer type (long) for the counter provides a much larger keyspace to prevent overflow for practical purposes.

The data structures for forward (long-to-short) and reverse (short-to-long) lookups become inconsistent.

How to Handle:

Both mappings must be updated atomically within the encode function to ensure that every encoded URL is decodable.

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:

If the `encode` method is called multiple times with the same long URL, should it consistently return the same short URL, or generate a new one each time?
What are the specific format requirements for the short URL? For instance, should it use the prefix `http://tinyurl.com/`, and are there any constraints on the length or character set (e.g., alphanumeric) for the unique code?
My encoding strategy must generate a unique code for each URL. If I use an approach like hashing or random generation, collisions are possible. How should my algorithm handle such collisions?
Is the mapping between long and short URLs expected to be persistent, or is it sufficient for the mapping to exist only for the lifetime of a single `Solution` object instance?
The problem guarantees that a `shortUrl` given to `decode` was previously generated. Does this also mean I can assume the format of the `shortUrl` string itself is always valid and doesn't require parsing validation?

Brute Force Solution

Approach

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

To shorten a long web address, start by creating a short, random sequence of characters.
Next, look through all the short links you've already created to see if this new random one has been used before.
If you find that it's already in use, you must throw it away and create another random sequence from scratch.
Keep repeating this process of creating a random sequence and checking if it's taken, over and over, until you finally make one that is completely new.
Once you have a unique new short link, record it alongside the original long web address so you can remember which is which.
When you need to find the original address from a short link, you just look up the short link in your records to find the long address you saved with it.

Code Implementation

class Codec:

    def __init__(self):
        self.url_database = []
        self.alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
        self.short_url_length = 6

    def encode(self, longUrl: str) -> str:
        while True:
            # Keep generating random names until we find one that is not already in our database.

            potential_short_name = "".join([random.choice(self.alphabet) for _ in range(self.short_url_length)])
            
            is_name_taken = False
            for _, existing_short_name in self.url_database:
                if existing_short_name == potential_short_name:
                    is_name_taken = True
                    break

            if not is_name_taken:
                # Once an unused short name is found, we record the pair in our 'notebook' for future lookups.

                self.url_database.append((longUrl, potential_short_name))
                
                return "http://tinyurl.com/" + potential_short_name

    def decode(self, shortUrl: str) -> str:
        short_code_to_find = shortUrl.replace("http://tinyurl.com/", "")
        
        # To find the original address, we must look through every entry one-by-one, just like a notebook.

        for original_address, short_code in self.url_database:
            
            if short_code == short_code_to_find:
                return original_address
        
        return ""

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Think of the system as having a simple counter, like a ticket dispenser, that starts at one and goes up.
When a new long web address comes in, we give it the next available number from our counter.
We then convert this number into a very compact code using a special character set that includes both letters and digits. This keeps the resulting code extremely short.
We store this connection: the original long address is now linked to this new, unique code.
Finally, we increment our counter, so it's ready with a fresh number for the next web address.
To get the long address back from a short code, we simply do the reverse. We translate the code back into its original number.
With that number, we can immediately look up and find the original long web address that we saved.

Code Implementation

class Codec:

    def __init__(self):
        self.alphabet_characters = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
        self.base_url_prefix = "http://tinyurl.com/"
        self.long_urls_list = []
        self.url_to_short_code_map = {}

    def encode(self, long_url: str) -> str:
        # To avoid creating duplicates, we first check if this URL already has a short version.

        if long_url in self.url_to_short_code_map:
            return self.url_to_short_code_map[long_url]

        # Storing the new URL gives it a unique, sequential ID based on its list position.

        self.long_urls_list.append(long_url)
        unique_id_number = len(self.long_urls_list) - 1

        # The ID is converted to a compact base-62 representation for the short URL code.

        short_identifier_code = ""
        if unique_id_number == 0:
            short_identifier_code = self.alphabet_characters[0]
        else:
            temporary_id_number = unique_id_number
            while temporary_id_number > 0:
                remainder_index = temporary_id_number % 62
                short_identifier_code = self.alphabet_characters[remainder_index] + short_identifier_code
                temporary_id_number //= 62
        
        full_short_url = self.base_url_prefix + short_identifier_code
        self.url_to_short_code_map[long_url] = full_short_url
        return full_short_url

    def decode(self, short_url: str) -> str:
        short_identifier_code = short_url.replace(self.base_url_prefix, "")
        unique_id_number = 0

        # The short code must be converted back to its original numeric ID to find the long URL.

        for character_in_code in short_identifier_code:
            unique_id_number = unique_id_number * 62 + self.alphabet_characters.find(character_in_code)

        # The decoded ID directly corresponds to the index of the original URL in our master list.

        return self.long_urls_list[unique_id_number]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
The encode method is called multiple times with the exact same longUrl.	The implementation should return the pre-existing shortUrl for that longUrl rather than creating a new one to ensure idempotency.
The decode method receives a full URL string, including the base path like 'http://tinyurl.com/', not just the unique code.	The solution must parse the input string to extract only the unique short code before performing the lookup.
The number of unique URLs to encode exhausts all possible combinations for a fixed-length short code.	A counter-based approach handles this by naturally increasing the length of the generated code as the counter value grows.
Two distinct longUrls map to the same shortUrl code, causing a collision.	A collision must be detected and resolved, for example by regenerating the code until a unique one is found.
The input longUrl is at the maximum allowed length of 10,000 characters.	The data structure, typically a hash map, must efficiently store and retrieve these large strings without significant performance degradation.
The encode method is called with a URL that is itself a previously generated short URL.	This input should be treated like any other valid URL and encoded to a new, different short URL.
An internal counter used for generating unique IDs exceeds the maximum value for its data type, like Integer.MAX_VALUE.	Using a 64-bit integer type (long) for the counter provides a much larger keyspace to prevent overflow for practical purposes.
The data structures for forward (long-to-short) and reverse (short-to-long) lookups become inconsistent.	Both mappings must be updated atomically within the encode function to ensure that every encoded URL is decodable.

The encode method is called multiple times with the exact same longUrl.

How to Handle:

The implementation should return the pre-existing shortUrl for that longUrl rather than creating a new one to ensure idempotency.

The decode method receives a full URL string, including the base path like 'http://tinyurl.com/', not just the unique code.

How to Handle:

The solution must parse the input string to extract only the unique short code before performing the lookup.

The number of unique URLs to encode exhausts all possible combinations for a fixed-length short code.

How to Handle:

A counter-based approach handles this by naturally increasing the length of the generated code as the counter value grows.

Two distinct longUrls map to the same shortUrl code, causing a collision.

How to Handle:

A collision must be detected and resolved, for example by regenerating the code until a unique one is found.

The input longUrl is at the maximum allowed length of 10,000 characters.

How to Handle:

The data structure, typically a hash map, must efficiently store and retrieve these large strings without significant performance degradation.

The encode method is called with a URL that is itself a previously generated short URL.

How to Handle:

This input should be treated like any other valid URL and encoded to a new, different short URL.

An internal counter used for generating unique IDs exceeds the maximum value for its data type, like Integer.MAX_VALUE.

How to Handle:

Using a 64-bit integer type (long) for the counter provides a much larger keyspace to prevent overflow for practical purposes.

The data structures for forward (long-to-short) and reverse (short-to-long) lookups become inconsistent.

How to Handle:

Both mappings must be updated atomically within the encode function to ensure that every encoded URL is decodable.

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