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

Problem Link

Description

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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.

Solution

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Python3

class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        res = set()
        
        def convertTime(time):
            return int("".join(time.split(':')))
        
        mp = collections.defaultdict(list)
        
        for name,time in zip(keyName, keyTime):
            mp[name].append(convertTime(time))
            
        for name in mp:
            time = sorted(mp[name])
            deq = collections.deque()
            c = 0
            
            for t in time:
                
                while deq and t - deq[0] > 100:
                    deq.popleft()
                    if c > 0:
                        c -= 1
                
                c += 1
                deq.append(t)
                
                if c >= 3:
                    res.add(name)
                    break
        
        return sorted(list(res))