Hash Map

Question 1

2933. High-Access Employees
      Medium
      Topics
      Companies: Atlassian
      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’.

https://leetcode.com/problems/high-access-employees/description

Answer 1

from itertools import starmap, groupby
from collections import deque,defaultdict
class Solution:
    def findHighAccessEmployees(self, access_times: List[List[str]]) -> List[str]:
        time_check = defaultdict(deque)
        def std_time(mil_time):
            mins = int(mil_time[2:])
            hrs = int(mil_time[:2])
            return mins + (hrs * 60)

        entries_sorted = sorted(starmap(lambda emp, t: (emp, std_time(t)), access_times), key=lambda x:(x[0], x[1]))
        hae = set()

        for emp, time_mins in entries_sorted:
            # We already marked this employe High Access, no need for further checks.
            if emp in hae:
                continue
            if emp not in time_check:
                time_check[emp].append(time_mins)
                continue
            else:
                while time_check[emp] and time_mins - time_check[emp][0] >= 60:
                    time_check[emp].popleft()
                time_check[emp].append(time_mins)
                if len(time_check[emp]) >= 3:
                    hae.add(emp)
        return list(hae)

O(nLog(n)), space O(n)