Leetcode #1797: Design Authentication Manager
In this guide, we solve Leetcode #1797 Design Authentication Manager 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
There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire timeToLive seconds after the currentTime.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Hash Table, Linked List, Doubly-Linked List
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
["AuthenticationManager", "renew", "generate", "countUnexpiredTokens", "generate", "renew", "renew", "countUnexpiredTokens"]
[[5], ["aaa", 1], ["aaa", 2], [6], ["bbb", 7], ["aaa", 8], ["bbb", 10], [15]]
Output
[null, null, null, 1, null, null, null, 0]
Explanation
AuthenticationManager authenticationManager = new AuthenticationManager(5); // Constructs the AuthenticationManager with timeToLive = 5 seconds.
authenticationManager.renew("aaa", 1); // No token exists with tokenId "aaa" at time 1, so nothing happens.
authenticationManager.generate("aaa", 2); // Generates a new token with tokenId "aaa" at time 2.
authenticationManager.countUnexpiredTokens(6); // The token with tokenId "aaa" is the only unexpired one at time 6, so return 1.
authenticationManager.generate("bbb", 7); // Generates a new token with tokenId "bbb" at time 7.
authenticationManager.renew("aaa", 8); // The token with tokenId "aaa" expired at time 7, and 8 >= 7, so at time 8 the renew request is ignored, and nothing happens.
authenticationManager.renew("bbb", 10); // The token with tokenId "bbb" is unexpired at time 10, so the renew request is fulfilled and now the token will expire at time 15.
authenticationManager.countUnexpiredTokens(15); // The token with tokenId "bbb" expires at time 15, and the token with tokenId "aaa" expired at time 7, so currently no token is unexpired, so return 0.
Python Solution
class AuthenticationManager:
def __init__(self, timeToLive: int):
self.t = timeToLive
self.d = defaultdict(int)
def generate(self, tokenId: str, currentTime: int) -> None:
self.d[tokenId] = currentTime + self.t
def renew(self, tokenId: str, currentTime: int) -> None:
if self.d[tokenId] <= currentTime:
return
self.d[tokenId] = currentTime + self.t
def countUnexpiredTokens(self, currentTime: int) -> int:
return sum(exp > currentTime for exp in self.d.values())
# Your AuthenticationManager object will be instantiated and called as such:
# obj = AuthenticationManager(timeToLive)
# obj.generate(tokenId,currentTime)
# obj.renew(tokenId,currentTime)
# param_3 = obj.countUnexpiredTokens(currentTime)
Complexity
The time complexity is O(n). The space complexity is , where is the number of key-value pairs in the hash table .
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.