LeetCode #1797 — MEDIUM

Design Authentication Manager

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode
The Problem

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. If the token is renewed, the expiry time will be extended to expire timeToLive seconds after the (potentially different) currentTime.

Implement the AuthenticationManager class:

  • AuthenticationManager(int timeToLive) constructs the AuthenticationManager and sets the timeToLive.
  • generate(string tokenId, int currentTime) generates a new token with the given tokenId at the given currentTime in seconds.
  • renew(string tokenId, int currentTime) renews the unexpired token with the given tokenId at the given currentTime in seconds. If there are no unexpired tokens with the given tokenId, the request is ignored, and nothing happens.
  • countUnexpiredTokens(int currentTime) returns the number of unexpired tokens at the given currentTime.

Note that if a token expires at time t, and another action happens on time t (renew or countUnexpiredTokens), the expiration takes place before the other actions.

Example 1:

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.

Constraints:

  • 1 <= timeToLive <= 108
  • 1 <= currentTime <= 108
  • 1 <= tokenId.length <= 5
  • tokenId consists only of lowercase letters.
  • All calls to generate will contain unique values of tokenId.
  • The values of currentTime across all the function calls will be strictly increasing.
  • At most 2000 calls will be made to all functions combined.
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. If the token is renewed, the expiry time will be extended to expire timeToLive seconds after the (potentially different) currentTime. Implement the AuthenticationManager class: AuthenticationManager(int timeToLive) constructs the AuthenticationManager and sets the timeToLive. generate(string tokenId, int currentTime) generates a new token with the given tokenId at the given currentTime in seconds. renew(string tokenId, int currentTime) renews the unexpired token with the given tokenId at the given currentTime in seconds. If there are no unexpired tokens with the given tokenId, the request is ignored, and nothing happens. countUnexpiredTokens(int currentTime) returns the number of unexpired

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Linked List · Design

Example 1

["AuthenticationManager","renew","generate","countUnexpiredTokens","generate","renew","renew","countUnexpiredTokens"]
[[5],["aaa",1],["aaa",2],[6],["bbb",7],["aaa",8],["bbb",10],[15]]
Step 02

Core Insight

What unlocks the optimal approach

  • Using a map, track the expiry times of the tokens.
  • When generating a new token, add it to the map with its expiry time.
  • When renewing a token, check if it's on the map and has not expired yet. If so, update its expiry time.
  • To count unexpired tokens, iterate on the map and check for each token if it's not expired yet.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1797: Design Authentication Manager
class AuthenticationManager {
    private int t;
    private Map<String, Integer> d = new HashMap<>();

    public AuthenticationManager(int timeToLive) {
        t = timeToLive;
    }

    public void generate(String tokenId, int currentTime) {
        d.put(tokenId, currentTime + t);
    }

    public void renew(String tokenId, int currentTime) {
        if (d.getOrDefault(tokenId, 0) <= currentTime) {
            return;
        }
        generate(tokenId, currentTime);
    }

    public int countUnexpiredTokens(int currentTime) {
        int ans = 0;
        for (int exp : d.values()) {
            if (exp > currentTime) {
                ++ans;
            }
        }
        return ans;
    }
}

/**
 * Your AuthenticationManager object will be instantiated and called as such:
 * AuthenticationManager obj = new AuthenticationManager(timeToLive);
 * obj.generate(tokenId,currentTime);
 * obj.renew(tokenId,currentTime);
 * int param_3 = obj.countUnexpiredTokens(currentTime);
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

COPY TO ARRAY
O(n) time
O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS
O(n) time
O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.