Encode And Decode Tinyurl Leetcode Problem 535 [Python Solution]

If you’ve ever used a URL shortening service like TinyURL, you’ve probably wondered how it works, Encode And Decode Tinyurl will teach.

In this article, we’ll explore the Encode And Decode Tinyurl problem, which is a common interview question and an interesting real-world application of data structures.

We’ll provide a Python solution to this problem and explain its time and space complexity.

So, let’s dive in!

Problem Overview

The problem statement for Encode And Decode Tinyurl can be summarized as follows:

You need to design a class that can encode a given URL into a shorter “tiny” URL and decode the tiny URL to retrieve the original URL.

The class should have two methods:

• encode(longUrl: str): This method takes a long URL as input and returns a shortened/tiny URL.
• decode(shortUrl: str): This method takes a tiny URL and returns the original long URL.

The key idea is to maintain a mapping between the original URL and the corresponding tiny URL, allowing for efficient encoding and decoding.

There are no strict restrictions on how the encoding and decoding should work, as long as the tiny URL can be decoded to the original URL.

Example 1:

Let’s illustrate this with an example:

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

Explanation:

obj = Solution()
tiny = obj.encode(url)  # Returns the encoded tiny URL.

ans = obj.decode(tiny)  # Returns the original URL after decoding it.

Constraints:

• The length of the URL is between 1 and 10^4.
• The input URL is guaranteed to be a valid URL.

To solve this problem efficiently, we can use a hash map to store the mapping between original URLs and their corresponding tiny URLs.

Understanding the Constraints

Before we dive into the solution, let’s discuss the constraints provided in the problem statement.

The primary constraint is the length of the URL, which is between 1 and 10^4 characters.

This means that the URLs can vary in length but are within a reasonable range for storage and processing.

As for the URL format, we are assured that the input URL is a valid URL.

Valid URLs adhere to a specific format, including the protocol (e.g., “http” or “https”), domain, and path.

This simplifies the problem since we don’t have to handle malformed or invalid URLs.

The problem doesn’t specify the number of URLs to be encoded and decoded, so our solution should be efficient for both individual URL operations and handling multiple URLs in succession.

Encode and Decode TinyURL Python Solution

Now, let’s take a look at the Python solution to this problem.

We’ll use a Codec class to implement the encoding and decoding methods.

class Codec:
    def __init__(self):
        self.encodeMap = {}
        self.decodeMap = {}
        self.base = "http://tinyurl.com/"

    def encode(self, longUrl: str) -> str:
        """Encodes a URL to a shortened URL."""
        if longUrl not in self.encodeMap: 
            shortUrl = self.base + str(len(self.encodeMap) + 1)
            self.encodeMap[longUrl] = shortUrl
            self.decodeMap[shortUrl] = longUrl
        return self.encodeMap[longUrl]

    def decode(self, shortUrl: str) -> str:
        """Decodes a shortened URL to its original URL."""
        return self.decodeMap[shortUrl]

The Codec class has the following components:

1. Constructor (__init__): In the constructor, we initialize three data members:

• encodeMap: A dictionary to map original URLs to their corresponding tiny URLs.
• decodeMap: A dictionary to map tiny URLs to their original URLs.
• base: A constant string representing the base of the tiny URLs (e.g., “http://tinyurl.com/”).

We’ll append a unique identifier to this base for each URL.

2. Encode (encode): This method encodes a long URL into a tiny URL.

Here’s how it works:

• First, we check if the long URL already exists in the encodeMap.

If it does, we return the corresponding tiny URL.

• If the long URL is not in the encodeMap, we generate a new tiny URL by appending a unique identifier to the base using the length of the encodeMap plus one as the identifier.
• We update both the encodeMap and decodeMap to maintain the mapping between the long URL and the tiny URL.
• Finally, we return the generated tiny URL.

1. Decode (decode): This method decodes a tiny URL into the original long URL.

We simply look up the tiny URL in the decodeMap and return the corresponding original URL.

The code is relatively straightforward, and it efficiently handles the encoding and decoding of URLs.

Time and Space Complexity

Let’s analyze the time and space complexity of our solution:

• Time Complexity:
• The encode function has an average time complexity of O(1) for encoding a URL.

This is because dictionary operations like checking for the existence of a key and adding a new key-value pair have an average time complexity of O(1).

• The decode function also has an average time complexity of O(1) for decoding a tiny URL since it involves a simple dictionary lookup.
• Space Complexity:
• The space complexity of our solution is influenced by the space required to store the mapping between URLs in the encodeMap and decodeMap.

In the worst case, where we encode a very large number of unique URLs, the space complexity is O(n), where n is the number of unique URLs encoded.

In practice, the space complexity is unlikely to be a concern, given the constraints of the problem.

Reasoning Behind Our Approach

The approach we’ve taken to solve the Encode And Decode Tinyurl problem is simple and efficient.

By using two dictionaries (encodeMap and decodeMap), we maintain a bidirectional mapping between original URLs and tiny URLs.

When we need to encode a long URL, we check if it’s already in the encodeMap.

If it is, we return the corresponding tiny URL.

If not, we generate a new tiny URL, update both maps, and return the tiny URL.

Decoding is as straightforward as looking up the tiny URL in the decodeMap.

The time complexity of our solution is excellent, with both encoding and decoding operations having an average time complexity of O(1).

The space complexity, while technically O(n) in the worst case, is unlikely to be a practical issue given the problem constraints.

In a real-world scenario, if you were implementing a URL shortening service, you might consider additional features and optimizations, such as handling custom short URLs, implementing expiration policies, and distributing the service across multiple servers.

However, for the purposes of the interview problem, our solution is both clear and efficient.

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Conclusion

In this article, we explored the Encode And Decode Tinyurl problem and provided a Python solution that efficiently encodes and decodes URLs.

We discussed the problem overview, the constraints we need to consider, and the reasoning

behind our approach.

Our solution leverages two dictionaries to maintain the mapping between long URLs and their corresponding tiny URLs, resulting in a straightforward and efficient solution.

If you’re preparing for coding interviews, it’s important to understand how to design data structures and algorithms for real-world scenarios like URL shortening.

Practice coding problems like this one to improve your problem-solving skills and ace your interviews.

Feel free to comment, ask questions, make suggestions, or share this content with others who might find it helpful.

You can find the original problem on LeetCode.

Happy coding!