Recursive Digit Sum problem hackerRank

 We define super digit of an integer  using the following rules:


Given an integer, we need to find the super digit of the integer.


If  has only  digit, then its super digit is .

Otherwise, the super digit of  is equal to the super digit of the sum of the digits of .

For example, the super digit of  will be calculated as:


super_digit(9875)    9+8+7+5 = 29 

super_digit(29) 2 + 9 = 11

super_digit(11) 1 + 1 = 2

super_digit(2) = 2  

Example



The number  is created by concatenating the string   times so the initial .


    superDigit(p) = superDigit(9875987598759875)

                  9+8+7+5+9+8+7+5+9+8+7+5+9+8+7+5 = 116

    superDigit(p) = superDigit(116)

                  1+1+6 = 8

    superDigit(p) = superDigit(8)

All of the digits of  sum to . The digits of  sum to .  is only one digit, so it is the super digit.


Function Description


Complete the function superDigit in the editor below. It must return the calculated super digit as an integer.


superDigit has the following parameter(s):


string n: a string representation of an integer

int k: the times to concatenate  to make 

Returns


int: the super digit of  repeated  times

Input Format


The first line contains two space separated integers


Solution :



import java.io.*;

class Result {

    /*
     * Complete the 'superDigit' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts the following parameters:
     *  1. STRING n
     *  2. INTEGER k
     */

    public static int superDigit(String n, int k) {
        long sum = 0;

        // Compute the sum of digits of n
        for (char digit : n.toCharArray()) {
            sum += digit - '0';  // Convert char to integer
        }

        // Multiply sum by k (since the number is repeated k times)
        sum *= k;

        // Compute the super digit using recursion
        return superDigitHelper(sum);
    }

    private static int superDigitHelper(long num) {
        if (num < 10) return (int) num; // Base case: if single-digit, return it

        long sum = 0;
        while (num > 0) {
            sum += num % 10;  // Extract last digit and add to sum
            num /= 10;         // Remove last digit
        }

        return superDigitHelper(sum); // Recur until a single-digit number is obtained
    }
}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String[] firstMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

        String n = firstMultipleInput[0];
        int k = Integer.parseInt(firstMultipleInput[1]);

        int result = Result.superDigit(n, k);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}




Comments

Post a Comment

Popular posts from this blog

Two Sum II - Input Array Is Sorted

 Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order , find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length. Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2. The tests are generated such that there is exactly one solution. You may not use the same element twice. Your solution must use only constant extra space .   Example 1: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2]. Example 2: Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3]. Example 3: Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1,...
Read more

Comparable Vs. Comparator in Java

  In Java , Comparable and Comparator both are interfaces used to sort objects. Comparable It defines  natural ordering  within a class. It is implemented  inside  the class being sorted. It uses the compareTo() method. It allows sorting  by one one . Comparable provides a  single sorting sequence . In other words, we can sort the collection on the basis of a single element such as id, name, and price. Comparable  affects the original class , i.e., the actual class is modified. Comparable is present in  java.lang  package. We can sort the list elements of Comparable type by  Collections.sort(List)  method. Only one compareTo() method can be implemented, restricting flexibility. Comparator It defines  custom sorting logic It is implemented  in a separate class  or using lambda expressions . It uses the compare() method. It allows sorting  by multiple criteria . The Comparator provides  multiple sorting...
Read more

Increasing Triplet Subsequence

 Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.   Example 1: Input: nums = [1,2,3,4,5] Output: true Explanation: Any triplet where i < j < k is valid. Example 2: Input: nums = [5,4,3,2,1] Output: false Explanation: No triplet exists. Example 3: Input: nums = [2,1,5,0,4,6] Output: true Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6. Brute Force : Failed for some testcase class Solution {     public boolean increasingTriplet ( int [] nums ) {         int n = nums . length ;         int i = 0 ;         int j =i+ 1 ;         int k =j+ 1 ;         while (i <n && j<n && k<n)         {         if (nums...
Read more