# Solving Leetcode Interviews in Seconds with AI: Alert Using Same Key-Card Three or More Times in a One Hour Period


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1604" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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. The system emits an alert if any worker uses the key-card three or more times in a one-hour period. You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person's name and the time when their key-card was used in a single day. Access times are given in the 24-hour time format "HH:MM", such as "23:51" and "09:49". Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically. Notice that "10:00" - "11:00" is considered to be within a one-hour period, while "22:51" - "23:52" is not considered to be within a one-hour period.   Example 1:  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").  Example 2:  Input: keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"] Output: ["bob"] Explanation: "bob" used the keycard 3 times in a one-hour period ("21:00","21:20", "21:30").    Constraints:  1 <= keyName.length, keyTime.length <= 105 keyName.length == keyTime.length keyTime[i] is in the format "HH:MM". [keyName[i], keyTime[i]] is unique. 1 <= keyName[i].length <= 10 keyName[i] contains only lowercase English letters.  

	# Explanation
	Here's a breakdown of the solution approach, followed by the Python code:

*   **Data Organization:** Group keycard usage times by worker name for efficient processing.
*   **Time Conversion and Sorting:** Convert the time strings "HH:MM" to minutes and sort the times in ascending order. This facilitates sliding window calculation.
*   **Sliding Window:** Use a sliding window to check if any worker has used the keycard three or more times within a one-hour (60-minute) period.

*   **Runtime Complexity:** O(N log N), where N is the number of keycard usages (due to sorting).
*   **Storage Complexity:** O(N) in the worst case, to store the keycard usage times for each worker.

	
	# Code
	```python
	def alert_names(keyName, keyTime):
    """
    Finds workers who used their keycards three or more times in a one-hour period.

    Args:
        keyName (list[str]): A list of worker names.
        keyTime (list[str]): A list of keycard access times in "HH:MM" format.

    Returns:
        list[str]: A sorted list of unique worker names who triggered an alert.
    """

    worker_times = {}
    for i in range(len(keyName)):
        name = keyName[i]
        time = keyTime[i]
        if name not in worker_times:
            worker_times[name] = []
        worker_times[name].append(time)

    alert_workers = set()
    for name, times in worker_times.items():
        minutes = []
        for time in times:
            hour, minute = map(int, time.split(':'))
            minutes.append(hour * 60 + minute)
        minutes.sort()

        for i in range(len(minutes) - 2):
            if minutes[i + 2] - minutes[i] <= 60:
                alert_workers.add(name)
                break  # No need to check further for this worker

    return sorted(list(alert_workers))
	```