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

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:

import bisect

class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        n = len(keyName)
        checkins = {}
        for i in range(n):
            name = keyName[i]
            time = keyTime[i]
            time = tuple(int(k) for k in time.split(":"))
            if name in checkins:
                bisect.insort(checkins[name], time)
            else:
                checkins[name] = [time]
        op = []
        names = sorted(checkins.keys())
        for name in names:
            m = len(checkins[name])
            alert = False
            for i in range(m):
                curr = checkins[name][i]
                j = bisect.bisect_right(checkins[name], (curr[0] + 1, curr[1]))
                if j - i >= 3:
                    alert = True
                    break
            if alert:
                op.append(name)
        return op