One Very Important User (VIU) has a Very Confidential Document (VCD) stored on his Dropbox account. He doesn't let anyone see the VCD, especially his roommates who often have access to his devices.
Opening the Dropbox mobile app on the VIU's tablet requires a four-digit passcode. To ensure the confidentiality of the stored information, the device is locked out of Dropbox after 10
consecutive failed passcode attempts. We need to implement a function that, given an array of attempts
made throughout the day and the correct passcode
, checks to see if the device should be locked, i.e. 10
or more consecutive failed passcode attempts were made.
Example
-
For
passcode = "1111"
andattempts = ["1111", "4444", "9999", "3333", "8888", "2222", "7777", "0000", "6666", "7285", "5555", "1111"]
the output should be
solution(passcode, attempts) = true
.The first attempt is correct, so the user must have successfully logged in. However, the next
10
consecutive attempts are incorrect, so the device should be locked. Thus, the output should betrue
. -
For
passcode = "1234"
andattempts = ["9999", "9999", "9999", "9999", "9999", "9999", "9999", "1234", "9999", "9999", "9999", "9999"]
the output should be
solution(passcode, attempts) = false
.There are only
9
consecutive incorrect attempts, so there's no need to lock the device.
Input/Output
-
[execution time limit] 4 seconds (py3)
-
[input] string passcode
String consisting of exactly
4
digits representing the correct passcode.Guaranteed constraints:
passcode.length = 4
. -
[input] array.string attempts
Array representing the passcode attempts in the order they were made. Each element of
attempts
is a string consisting of exactly4
digits.Guaranteed constraints:
0 ≤ attempts.length ≤ 20
,
attempts[i].length = 4
. -
[output] boolean
true
if10
or more consecutive failed passcode attempts were made,false
otherwise.
[Python 3] Syntax Tips
# Prints help message to the console
# Returns a string
def helloWorld(name):
print("This prints to the console when you Run Tests")
return "Hello, " + name
def solution(passcode, attempts):
aux = 0
for i in range(0,len(attempts)):
if attempts[i] == passcode:
aux = 0
else:
aux += 1
if aux == 10:
return True
return False