Leetcode #535: Encode and Decode TinyURL
In this guide, we solve Leetcode #535 Encode and Decode TinyURL in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Hash Table, String, Hash Function
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Codec:
def __init__(self):
self.m = defaultdict()
self.idx = 0
self.domain = 'https://tinyurl.com/'
def encode(self, longUrl: str) -> str:
"""Encodes a URL to a shortened URL."""
self.idx += 1
self.m[str(self.idx)] = longUrl
return f'{self.domain}{self.idx}'
def decode(self, shortUrl: str) -> str:
"""Decodes a shortened URL to its original URL."""
idx = shortUrl.split('/')[-1]
return self.m[idx]
# Your Codec object will be instantiated and called as such:
# codec = Codec()
# codec.decode(codec.encode(url))
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.