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
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