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  1. Recursive Digit Sum
  2. Discussions

Recursive Digit Sum

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206 Discussions

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  • ApolionXD
     + 0 comments

    Java:

    I dont know what im doing wrong:

    public static int superDigit(String n, int k) {
        StringBuilder superNBuilder = new StringBuilder(n);
        for(int i=0;i<k-1;i++)
        {
            superNBuilder.append(n);
        }
        String superN = superNBuilder.toString();
        BigInteger result = new BigInteger(""+superN.charAt(0));
        while(superN.length()>1){
            result=BigInteger.ZERO;
            for(int i=0;i<superN.length();i++){
                result=result.add(new BigInteger(""+superN.charAt(i)));
            }
            superN=result.toString();
        }
        return result.intValue();
    }

  • stanczello
     + 1 comment

    I don't get it. Described example '9875 4' results with super digit 2, but in Test case 1 with the same input the expected output is 8...

  • emailthegyro
     + 0 comments

    C++ solution

    int superDigit(string n, int k) {
    long long sum = 0;
    for(char chr : n)
        sum += chr - '0';
    sum *=k;
    while (sum>9){
        long long sum2 =0;
        while (sum > 0){
            sum2 += sum % 10;
            sum = sum/10;
        }
        sum = sum2;
        
    }     
    return sum; 
}

  • trupti29
     + 0 comments

    def superDigit(n, k):

    str1 = str(n)
sum1 =  0

for i in str1:
    sum1 += int(i)

sum1 = sum1*k
str2 = str(sum1)

while len(str2) != 1:
    sum2 = 0
    for i in str2:
        sum2 += int(i)
    str2 = str(sum2)
return sum2

  • valiruziboyev212
     + 0 comments
    Easy JavaScript Solution 
    function superDigit(n, k) {
    if(n.length === 1){
        return n
    }
    let p = 0
    for(const num of n){
        const z = k*Number(num)
        p+=z
    }
   return superDigit(p.toString(), 1)
}