Introduction

In this blog post, we will explore how to solve the LeetCode problem "2933" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day. The access time is represented as four digits using a 24-hour time format, for example, "0800" or "2250". An employee is said to be high-access if he has accessed the system three or more times within a one-hour period. Times with exactly one hour of difference are not considered part of the same one-hour period. For example, "0815" and "0915" are not part of the same one-hour period. Access times at the start and end of the day are not counted within the same one-hour period. For example, "0005" and "2350" are not part of the same one-hour period. Return a list that contains the names of high-access employees with any order you want. Example 1: Input: access_times = [["a","0549"],["b","0457"],["a","0532"],["a","0621"],["b","0540"]] Output: ["a"] Explanation: "a" has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21. But "b" does not have more than two access times at all. So the answer is ["a"]. Example 2: Input: access_times = [["d","0002"],["c","0808"],["c","0829"],["e","0215"],["d","1508"],["d","1444"],["d","1410"],["c","0809"]] Output: ["c","d"] Explanation: "c" has three access times in the one-hour period of [08:08, 09:07] which are 08:08, 08:09, and 08:29. "d" has also three access times in the one-hour period of [14:10, 15:09] which are 14:10, 14:44, and 15:08. However, "e" has just one access time, so it can not be in the answer and the final answer is ["c","d"]. Example 3: Input: access_times = [["cd","1025"],["ab","1025"],["cd","1046"],["cd","1055"],["ab","1124"],["ab","1120"]] Output: ["ab","cd"] Explanation: "ab" has three access times in the one-hour period of [10:25, 11:24] which are 10:25, 11:20, and 11:24. "cd" has also three access times in the one-hour period of [10:25, 11:24] which are 10:25, 10:46, and 10:55. So the answer is ["ab","cd"]. Constraints: 1 <= access_times.length <= 100 access_times[i].length == 2 1 <= access_times[i][0].length <= 10 access_times[i][0] consists only of English small letters. access_times[i][1].length == 4 access_times[i][1] is in 24-hour time format. access_times[i][1] consists only of '0' to '9'.

Explanation

Here's the solution to the high-access employees problem:

• Data Grouping and Sorting: Group access times by employee name and then sort the access times for each employee in ascending order. This ensures efficient checking of consecutive access times.
• Sliding Window: Iterate through the sorted access times for each employee using a sliding window of size 3. For each window, check if the difference between the first and last access time is less than 60 minutes.

• High-Access Identification: If the difference is less than 60 minutes, the employee is marked as high-access, and their name is added to the result list if it's not already present.

• Runtime Complexity: O(N log N), where N is the number of access time entries, dominated by sorting.

• Storage Complexity: O(N) in the worst case, where N is the number of unique employee names, for storing the access_times_by_employee dictionary and result list.

Code

    def findHighAccessEmployees(access_times):
    access_times_by_employee = {}
    for employee, time in access_times:
        if employee not in access_times_by_employee:
            access_times_by_employee[employee] = []
        access_times_by_employee[employee].append(time)

    high_access_employees = []
    for employee, times in access_times_by_employee.items():
        times.sort()
        for i in range(len(times) - 2):
            time1 = int(times[i])
            time2 = int(times[i + 2])

            hour1 = time1 // 100
            minute1 = time1 % 100
            hour2 = time2 // 100
            minute2 = time2 % 100

            diff = (hour2 * 60 + minute2) - (hour1 * 60 + minute1)

            if diff < 60 and employee not in high_access_employees:
                high_access_employees.append(employee)

    return high_access_employees