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

CamelCase

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1914 Discussions

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  • SVVBathini_w
     + 0 comments

    For Python3 Platform

    I wrote the code from scratch just to get more practice

    def camelcase(s):
    result = 1
    
    for i in s:
        if(i.isupper()):
            result = result + 1
    
    return result

s = input()

result = camelcase(s)

print(result)

  • 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