Leetcode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period

In this guide, we solve Leetcode #1604 Alert Using Same Key-Card Three or More Times in a One Hour Period 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.

Leetcode

Problem Statement

LeetCode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker's name and the time when it was used.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Array, Hash Table, String, Sorting

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: keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
Output: ["daniel"]
Explanation: "daniel" used the keycard 3 times in a one-hour period ("10:00","10:40", "11:00").

Python Solution

class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        d = defaultdict(list)
        for name, t in zip(keyName, keyTime):
            t = int(t[:2]) * 60 + int(t[3:])
            d[name].append(t)
        ans = []
        for name, ts in d.items():
            if (n := len(ts)) > 2:
                ts.sort()
                for i in range(n - 2):
                    if ts[i + 2] - ts[i] <= 60:
                        ans.append(name)
                        break
        ans.sort()
        return ans

Complexity

The time complexity is $O(n \times \log n)$ , and the space 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.

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.

Python Solution

class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        d = defaultdict(list)
        for name, t in zip(keyName, keyTime):
            t = int(t[:2]) * 60 + int(t[3:])
            d[name].append(t)
        ans = []
        for name, ts in d.items():
            if (n := len(ts)) > 2:
                ts.sort()
                for i in range(n - 2):
                    if ts[i + 2] - ts[i] <= 60:
                        ans.append(name)
                        break
        ans.sort()
        return ans

Leetcode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #1604: Alert Using Same Key-Card Three or More Times in a One Hour Period

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview