jay_mistry456 def minimumNumber(n, password): count = 0 if any(i.isdigit() for i in password)==False: count+=1 if any(i.islower() for i in password)==False: count+=1 if any(i.isupper() for i in password)==False: count+=1 if any(i in '!@#$%^&*()-+' for i in password)==False: count+=1 return max(count,6-n)
arunmvthemaster nice logic. Loved it!
mattopie [deleted]
AffineStructure Instead of using ==False, you can use not any
sayre_precure [deleted]sayre_precure Got one pretty similar :)
count = 0 cases = [r'[a-z]', r'[A-Z]', r'[\d]', r'[!@#$%^&*()\-+]'] for case in cases: if not re.search(case, password): count += 1 return max(count, 6 - n)
srtk1996 [deleted]
saro_mano Why does '-' alone needs a escape sequence ?
KamalTudu [deleted]t_mate_laszlo "This pattern type matches any one of the characters in the brackets, including escape sequences. Special characters like the dot(.) and asterisk (*) are not special inside a character set, so they don't need to be escaped. You can specify a range of characters by using a hyphen, as the following examples illustrate." - MDN (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions)
However [a-z] means characters from a to z so hiphen has to be escaped.
mukeshsharma1201 you can put - as first character in set, to automatically escape it.
Mike_Williams This is a great solution!
You can simplify it further by changing
[!@#$%^&*()-+] to \W+ which is the regex for Any Non-alphanumeric character
So you would have...
count = 0 cases = [r'[a-z]', r'[A-Z]', r'[\d]', r'\W+'] for case in cases: if not re.search(case, password): count += 1 return max(count, 6 - n)
ctrs1 Not an appropriate substitute, even if the test cases don't catch it.
\W+includes characters?,<,>, etc. which are not valid special characters based on the prompt.
venkat_san19994 can u please explain what that "if not " will do
DsR90 what is the time complexity for this...
mike_buttery Four checks O(n) plus final check O(1) = O(n)
bhanudtg why did you use max(count,6-n) can't i print the count value directly??
kj636441 please explain max()
mike_buttery The minimum length of the password is 6 characters and the minimum conditional characters is 4. max() ensures that the answer complies with both requirements.
kris_tobiasson I went full Regex in JS with this solution:
if(n <= 3){ return 6 - n; } let count = 0; if(password.match(/\d/) === null){ count++; } if(password.match(/[A-Z]/) === null){ count++; } if(password.match(/[a-z]/) === null){ count++; } if(password.match(/\W/) === null){ count++; } return (count + n >= 6) ? count : 6 - n;
ahmad1_adnan95 thanx man
deb1924 RegEx makes life easier! thanks for the soln
AlbertCahyawan should'nt 'n<= 2 ' since we only return the differences if more than those 4 other as 6 - 4 is 2 and prob it only take care for case there is need 5 more character as it there not empty input of password unless there is that test case just in case
aravindan7822 Why did you include max atlast ??
kris_tobiasson max() is a function in python that returns the largest of 2 or more items.
24god10 because "count" and "6 - n" (number of character need to add in string to full length 6 if n less than 6) both is number character to adding into string to satisfy, "count" will cover "6 - n" if "6 - n" less than "count", if we adding "count", we also satisfied "6 - n". And "6 - n" must be priority than "count" to satisfy "length must at least 6" in precondition.
yashzawar2000 great approach!
ilpropheta C++ folks:
bitsetis your friend here:int CharFamily(char c) { if(c >= '0' && c <= '9') return 0; if(c >= 'a' && c <= 'z') return 1; if(c >= 'A' && c <= 'Z') return 2; return 3; } int main() { int n; cin >> n; char c; bitset<4> mask; while (cin >> c) { mask.set(CharFamily(c)); } cout << max(6-n, 4-(int)mask.count()); }
facitoo nice logic!
eswarsai00009 but here if the special characters are different from the given then also we would return 3. so i think its a wrong logic
ilpropheta You've just found a tradeoff: the logic fits nothing but the problem requirements. If you need more fine-grained control over the special characters then you have to change the algorithm a bit.
harshhgupta16 how the length>6 condition is getting checked
ilpropheta It's handled here:
cout << max(6-n, 4-(int)mask.count())
Since we can only add characters - we cannot change any - the number of extra characters to add is the maximum between:
6-n(aka: number of characters missing to get length 6)4-mask.count()(aka: number of characters missing to get all the 4 required "families").
eNgiNeeR_giRl excellent.
quancntt05 This is my Solution (Java 8)
private static int checkPass(String s) { int count = 0; Pattern patternDigit = Pattern.compile("(\\d)"); Pattern patternUpper = Pattern.compile("([A-Z])"); Pattern patternLower = Pattern.compile("([a-z])"); Pattern patternSpecial = Pattern.compile("(\\W)"); Matcher matcherDigit = patternDigit.matcher(s); Matcher matcherUpper = patternUpper.matcher(s); Matcher matcherLower = patternLower.matcher(s); Matcher matcherSpecial = patternSpecial.matcher(s); if (!matcherDigit.find()) { count++; } if (!matcherUpper.find()) { count++; } if (!matcherLower.find()) { count++; } if (!matcherSpecial.find()) { count++; } if ((count+s.length())<6) { count = count + 6-(count+s.length()); } return count; }
vaibhav94vijay The last condition also works in this way, and thinks it is the small one too.
if ((count+str.length())<6) { count = 6-str.length(); }
Is It????
harpreet115 The question mentioned special characters as only : !@#$%^&()-+
\W will match all special characters. You ned to change the regex.
gabrielbb0306 You don't need to pre-compile regex expressions if you are going to use them once. This is my solution:
static int minimumNumber(int n, String password) {
int count = 0; if(!password.matches(".*[a-z].*")) { count++; } if(!password.matches(".*[A-Z].*")) { count++; } if(!password.matches(".*[0-9].*")) { count++; } if(password.matches("[a-zA-Z0-9]*")) { count++; } int lengthDifference = 6 - password.length(); if(lengthDifference > 0 && count <= lengthDifference) { return lengthDifference; } return count; }shellymane Can you please explain why you did this:
if(password.matches("[a-zA-Z0-9]*")) { count++; }
pawansinghla it escape a-z ,A-Z and 0-9 means only considered all special character.
souravp_995 ^[a-zA-Z0-9]* should be there.
petprog A better explanation of your code by ignoring the last if statement
return Math.max(count, lengthDifference);
matharumanpreet what can be more concise than this piece of JS, don't mind the syntax highlighting
function minNumber(n, pass) { const c = [/[0-9]/, /[a-z]/, /[A-Z]/, /[!@#$%^&*()\-+"]/] .map(r => !r.test(pass)) .filter(Boolean).length return Math.max(c,6-n); }
rahulbhadhoriya1 care to explain a little bit?
matharumanpreet The array is basically a list of regex's that need to be satisfied for a strong password and then we just test the password for each using array map and get back an array of booleans and then we filter out the truthy values and count and return the number of modifications to the original string
vamsikrishnared9 Thank you so much
gartenkralle Loled hard as I saw the amount of test cases. :D
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