2933. High-Access Employees

Problem Link

Description

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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'.

Solution

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Python3

class Solution:
    def findHighAccessEmployees(self, access_times: List[List[str]]) -> List[str]:
        emp = defaultdict(list)
        res = []
        
        for e, t in access_times:
            emp[e].append(int(t))
        
        for e in emp.keys():
            N = len(emp[e])
            if N < 3: continue
                
            emp[e].sort()
            V = emp[e]
            
            for a, b, c in zip(V, V[1:], V[2:]):
                if c - a < 100:
                    res.append(e)
                    break
        
        return res