Python one-liner with recursion:

return n if k == 1 and n < 10 else superDigit(k * sum(int(d) for d in list(str(n))), 1)

Typescript Solution

   if (n.length === 1) {
        return +n;
    }
    let sum: number = n.split('')
    .reduce((a, b) => { 
        return +a + +b 
    }, 0);    
    sum *= k;
    
    return superDigit(sum.toString(), 1) ;

def superDigit(n, k): # Calculate the initial sum for the repeated number num = str(sum(map(int, str(n))) * k)

# Define the helper function for finding the super digit
while len(num) > 1:
    num = str(sum(map(int, num)))

return int(num)

`//Java solution with recursion:-

public static int superDigit(String n, int k) { if(n.length()==1) return Integer.parseInt(n);

long sum = 0;
int i =0;
int[] digitArr = new int[n.length()];

while(i<n.length()){
    digitArr[i]=Integer.parseInt(n.charAt(i)+"");
    i++;
}

for(int j=0; j<digitArr.length;j++){
    if(digitArr[j]!=0)
        sum+=digitArr[j]*k;
}
//Calling the function recursively with n = String value of sum 
// and k = 1 as there is no repition required
return superDigit(String.valueOf(sum), 1); 
}

`

There is no recursion here, checkout my 3 line solution:

public static int superDigit(String n, int k) {
        BigInteger num=new BigInteger(n);
        int rem=  num.remainder(new BigInteger("9")).multiply(new BigInteger(k+"")).intValue();
   
    return rem%9==0?9: rem%9;
    }