LeetCode 535. Encode and Decode TinyURL



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"

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


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


For encoding, we can convert the long URL to hash code value and store it into a map. For decoding, we just need to use the hash code value from the short URL to get the original long URL.

Python Solution

class Codec:
    def __init__(self):
        self.map = {}
    def encode(self, longUrl: str) -> str:
        """Encodes a URL to a shortened URL.
        self.map[str(hash(longUrl))] = longUrl
        return "http://tinyurl.com/" + str(hash(longUrl));

    def decode(self, shortUrl: str) -> str:
        """Decodes a shortened URL to its original URL.
        shortUrl_hash = shortUrl.split("http://tinyurl.com/")[1]
        return self.map[shortUrl_hash]   

# Your Codec object will be instantiated and called as such:
# codec = Codec()
# codec.decode(codec.encode(url))
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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