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  1. Prepare
  2. Algorithms
  3. Strings
  4. CamelCase
  5. Discussions

CamelCase

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1913 Discussions

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  • leonardofa
     + 0 comments

    in javascript the solution for me was just a line:

    return s.split(/[A-Z]/).length

  • lrmeena7878
     + 0 comments

    def camelcase(s): c = 1 for i in range(len(s)): if s[i] == s[i].upper(): c += 1 return c

  • mr7323545
     + 0 comments

    My solution in java 15

    var sarr= s.split("(?=\p{Lu})"); return sarr.length;

  • jonathan_aceves1
     + 0 comments
    def minimumNumber(n, password):
    # Return the minimum number of characters to make the password strong
    minLength = 6
    patterns = [r'[0-9]', r'[a-z]', r'[A-Z]', r'[!@#$%^&\*\(\)\-\+]']
    total_chars = sum([1 for pattern in patterns if not re.search(pattern, password)])
    total_chars += (0 if minLength <= n + total_chars else minLength - n - total_chars)
    return total_chars

  • gopikan299
     + 0 comments

    my python solution

    def camelcase(s):

    tot = 1
for char in s:
    if char.isupper():
        tot += 1
return tot