Reduce string by performing repeated digit sum in group of K

Given a string S of length N consisting of digits and an integer K, Reduce the string by performing the following operation till the length of the string is greater than K:

1. Divide the string into consecutive groups of size K such that the first K characters are in the first group, the next K characters are in the second group, and so on. Note that the size of the last group can be smaller than K.
2. Merge the digit sum of consecutive groups together to form a new string. If the length of the string is greater than K, repeat the steps.

Examples:

Input: S = “11111222223”, K = 3
Output: “135”
Explanation: 
For the first round, divide S into groups of size 3: “111”, “112”, “222”, and “23”. 
Then calculate the digit sum of each group: 
1 + 1 + 1 = 3, 1 + 1 + 2 = 4,  2 + 2 + 2 = 6, and 2 + 3 = 5. 
So, string becomes “3” + “4” + “6” + “5” = “3465”. 
For the second round, divide s into “346” and “5”. 
Then calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
So, string becomes “13” + “5” = “135”. 
After second round. Now, length of S <= K, so return “135” as the answer.

Input: S = “1123”, K = 2
Output: “25”
Explanation: For the first round, divide S into groups of size 2: “11”, “23”. ​
Then we calculate the digit sum of each group: 1 + 1 = 2 and 2 + 3 = 5. 
So, String becomes “2” + “5” = “25” after the first round. 
Now, length of S <= K, so return “25” as the answer.

Approach: This is a simple implementation based problem. The solution is as follows:

Iterate S until its size is greater than K. In every iteration, divide the string in consecutive K sized groups, get their sum and add them to form the new string.

Follow the below steps to implement the idea:

• Run a loop until the given string S has length greater than K.
  • Take one new string (say temp) to store the new string. 
  • Traverse string S from i = 0 to i = s.size() – 1.
    • Define a variable (say sum = 0) to store the sum. 
    • Iterate in a group of K characters and keep storing the digit sum of K digits.
    • Append the sum in temp string.
  • Make S = temp, i.e. same as the newly formed string.
• Return S as the final string.

Below is the implementation for the above approach:

C++

// C++ code to implement the approach\

#include <bits/stdc++.h>
using namespace std;

// Function for finding the sum of Digits
// of number in group of k
string digitSum(string s, int k)
{
    // Run the loop until given string s
    // length is greater than k otherwise
    // return the string
    while (s.size() > k) {

        // Temp variable to store new string
        string temp = "";
        for (int i = 0; i < s.size(); i++) {
            int sum = 0;
            int count = 0;

            // Run the loop until count is
            // less than k or iterator is
            // less than the size of string
            while (i < s.size() && count < k) {

                // Sum up the string value
                sum += s[i++] - '0';

                // Increase the count
                count++;
            }

            // Add the obtained sum to the
            // temp variable
            temp += to_string(sum);
            i--;
        }

        // Update the value of main string
        s = temp;
    }

    // Return the final string
    return s;
}

// Driver code
int main()
{
    string S = "11111222223";
    int K = 3;

    // Function call
    string ans = digitSum(S, K);
    cout << ans << endl;
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG
{

  // Function for finding the sum of Digits
  // of number in group of k
  public static String digitSum(String s, int k)
  {

    // Run the loop until given string s
    // length is greater than k otherwise
    // return the string
    while (s.length() > k) {

      // Temp variable to store new string
      String temp = "";
      for (int i = 0; i < s.length(); i++) {
        int sum = 0;
        int count = 0;

        // Run the loop until count is
        // less than k or iterator is
        // less than the size of string
        while (i < s.length() && count < k) {

          // Sum up the string value
          sum += s.charAt(i++) - '0';

          // Increase the count
          count++;
        }

        // Add the obtained sum to the
        // temp variable
        temp += Integer.toString(sum);
        i--;
      }

      // Update the value of main string
      s = temp;
    }

    // Return the final string
    return s;
  }

  // Driver Code
  public static void main(String[] args)
  {
    String S = "11111222223";
    int K = 3;

    // Function call
    String ans = digitSum(S, K);
    System.out.println(ans);
  }
}

// This code is contributed by Rohit Pradhan

Python3

# Python code to implement the approach

# Function for finding the sum of Digits
# of number in group of k
def digitSum(s,k):

    # Run the loop until given string s
    # length is greater than k otherwise
    # return the string
    while(len(s)>k):

        # Temp variable to store new string
        temp=""
        # using while loop instead of for loop
        # because there is need to change value
        # of i which is not possible in python
        # in for loop
        i=0
        while(i<len(s)):
            Sum=0
            count=0

            # Run the loop until count is
            # less than k or iterator is
            # less than the size of string
            while(i<len(s) and count<k):
                # Sum up the string value
                Sum = Sum + int(s[i])
                i=i+1

                # Increase the count
                count=count+1

            # Add the obtained sum to the
            # temp variable

            temp = temp+str(Sum)
            i=i-1
            # Increment to make while loop Run
            # like for loop
            i=i+1

        # Update the value of main string
        s=temp
    # Return the final string
    return s

# Driver code
S = "11111222223"
K = 3

# Function call
ans = digitSum(S,K)
print(ans)

# This code is contributed by Pushpesh Raj

C#

// C# program to implement
// the above approach
using System;
class GFG
{
  // Function for finding the sum of Digits
  // of number in group of k
  public static string digitSum(string s, int k)
  {

    // Run the loop until given string s
    // length is greater than k otherwise
    // return the string
    while (s.Length > k) {

      // Temp variable to store new string
      string temp = "";
      for (int i = 0; i < s.Length; i++) {
        int sum = 0;
        int count = 0;

        // Run the loop until count is
        // less than k or iterator is
        // less than the size of string
        while (i < s.Length && count < k) {

          // Sum up the string value
          sum += s[i++] - '0';

          // Increase the count
          count++;
        }

        // Add the obtained sum to the
        // temp variable
        temp += sum;
        i--;
      }

      // Update the value of main string
      s = temp;
    }

    // Return the final string
    return s;
  }

// Driver Code
public static void Main()
{
    string S = "11111222223";
    int K = 3;

    // Function call
    String ans = digitSum(S, K);
    Console.Write(ans);
}
}

// This code is contributed by code_hunt.

Javascript

<script>
// Javascript code to implement the approach\

// Function for finding the sum of Digits
// of number in group of k
function digitSum(s, k)
{

    // Run the loop until given string s
    // length is greater than k otherwise
    // return the string
    while (s.length > k) {

        // Temp variable to store new string
        let temp = "";
        for (let i = 0; i < s.length; i++) {
            let sum = 0;
            let count = 0;

            // Run the loop until count is
            // less than k or iterator is
            // less than the size of string
            while (i < s.length && count < k) {

                // Sum up the string value
                sum += s[i++] - '0';

                // Increase the count
                count++;
            }

            // Add the obtained sum to the
            // temp variable
            temp += sum.toString();
            i--;
        }

        // Update the value of main string
        s = temp;
    }

    // Return the final string
    return s;
}

// Driver code
let S = "11111222223";
let K = 3;

// Function call
let ans = digitSum(S, K);
document.write(ans);

// This code is contributed by Samim Hossain Mondal.
</script>

Output

135

Time Complexity: O(N * K)
Auxiliary Space: O(N)