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My solution in Java language. After solved this problem, i believe there is no need to use recursion.

publicstaticintsuperDigit(Stringn,intk){// Write your code here/* Realizes that the super_digit of x is the sum of digits of x moduled by 9. Then realizes that value is equal to remain of x when divided by 9. However, the string is too long :( Sum of digits of that string is under 9*100000(according to the constraints) Edge case: super_digit(x) = 0 <=> x = 0 Steps: 1. Calculating the sum of digits. 2. Multipling the result with k, then calculating the result % 9 3. Check if the result is 0 and the sum is not 0 -> then return 9 Else return that result. Time complexity: O(L) with L is the length of the string Space complexity: O(1) This approach is still able to be better */intsum=0;//SUMfor(inti=0;i<n.length();i++){sum+=n.charAt(i)-'0';}intresult=sum;result%=9;//Just to ensure no error with long string and big kresult*=k;//K timesresult%=9;//The remain after divide by 9return(result==0&&sum!=0)?9:result;}

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## Recursive Digit Sum

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My solution in Java language. After solved this problem, i believe there is no need to use recursion.