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Given a number as a string, find the number of contiguous subsequences which recursively add up to 9
  • Difficulty Level : Easy
  • Last Updated : 08 Jan, 2019

Given a number as a string, write a function to find the number of substrings (or contiguous subsequences) of the given string which recursively add up to 9.

For example digits of 729 recursively add to 9,
7 + 2 + 9 = 18
Recur for 18
1 + 8 = 9

Examples:

 
Input: 4189
Output: 3
There are three substrings which recursively add to 9.
The substrings are 18, 9 and 189.

Input: 999
Output: 6
There are 6 substrings which recursively add to 9.
9, 99, 999, 9, 99, 9

All digits of a number recursively add up to 9, if only if the number is multiple of 9. We basically need to check for s%9 for all substrings s. One trick used in below program is to do modular arithmetic to avoid overflow for big strings.

Following is a simple implementation based on this approach. The implementation assumes that there are no leading 0’s in input number.

C++






// C++ program to count substrings with recursive sum equal to 9
#include <iostream>
#include <cstring>
using namespace std;
  
int count9s(char number[])
{
    int count = 0; // To store result
    int n = strlen(number);
  
    // Consider every character as beginning of substring
    for (int i = 0; i < n; i++)
    {
        int sum = number[i] - '0'//sum of digits in current substring
  
        if (number[i] == '9') count++;
  
        // One by one choose every character as an ending character
        for (int j = i+1; j < n; j++)
        {
            // Add current digit to sum, if sum becomes multiple of 5
            // then increment count. Let us do modular arithmetic to
            // avoid overflow for big strings
            sum = (sum + number[j] - '0')%9;
  
            if (sum == 0)
               count++;
        }
    }
    return count;
}
  
// driver program to test above function
int main()
{
    cout << count9s("4189") << endl;
    cout << count9s("1809");
    return 0;
}


Java




// Java program to count
// substrings with 
// recursive sum equal to 9
import java.io.*;
  
class GFG
{
static int count9s(String number)
{
    // To store result
    int count = 0
    int n = number.length();
  
    // Consider every character 
    // as beginning of substring
    for (int i = 0; i < n; i++)
    {
        // sum of digits in
        // current substring
        int sum = number.charAt(i) - '0'
  
        if (number.charAt(i) == '9'
            count++;
  
        // One by one choose 
        // every character as 
        // an ending character
        for (int j = i + 1;
                 j < n; j++)
        {
            // Add current digit to 
            // sum, if sum becomes 
            // multiple of 5 then 
            // increment count. Let
            // us do modular arithmetic 
            // to avoid overflow for 
            // big strings
            sum = (sum +
                   number.charAt(j) - 
                            '0') % 9;
  
            if (sum == 0)
            count++;
        }
    }
    return count;
}
  
// Driver Code
public static void main (String[] args) 
{
    System.out.println(count9s("4189"));
    System.out.println(count9s("1809"));
}
}
  
// This code is contributed 
// by anuj_67.


Python 3




# Python 3 program to count substrings
# with recursive sum equal to 9
  
def count9s(number):
  
    count = 0 # To store result
    n = len(number)
  
    # Consider every character as 
    # beginning of substring
    for i in range(n):
          
        # sum of digits in current substring
        sum = ord(number[i]) - ord('0')     
  
        if (number[i] == '9'):
            count += 1
  
        # One by one choose every character 
        # as an ending character
        for j in range(i + 1, n):
          
            # Add current digit to sum, if 
            # sum becomes multiple of 5 then
            # increment count. Let us do 
            # modular arithmetic to avoid 
            # overflow for big strings
            sum = (sum + ord(number[j]) - 
                         ord('0')) % 9
  
            if (sum == 0):
                count += 1
    return count
  
# Driver Code
if __name__ == "__main__":
      
    print(count9s("4189"))
    print(count9s("1809"))
  
# This code is contributed by ita_c


C#




// C# program to count
// substrings with 
// recursive sum equal to 9
using System;
class GFG
{
static int count9s(String number)
{
    // To store result
    int count = 0; 
    int n = number.Length;
  
    // Consider every character 
    // as beginning of substring
    for (int i = 0; i < n; i++)
    {
        // sum of digits in
        // current substring
        int sum = number[i] - '0'
  
        if (number[i] == '9'
            count++;
  
        // One by one choose 
        // every character as 
        // an ending character
        for (int j = i + 1;
                 j < n; j++)
        {
            // Add current digit to 
            // sum, if sum becomes 
            // multiple of 5 then 
            // increment count. Let
            // us do modular arithmetic 
            // to avoid overflow for 
            // big strings
            sum = (sum + number[j] - 
                           '0') % 9;
  
            if (sum == 0)
            count++;
        }
    }
    return count;
}
  
// Driver Code
public static void Main () 
{
    Console.WriteLine(count9s("4189"));
    Console.WriteLine(count9s("1809"));
}
}
  
// This code is contributed 
// by anuj_67.


PHP




<?php
// PHP program to count substrings
// with recursive sum equal to 9
  
function count9s($number)
{
    // To store result
    $count = 0; 
    $n = strlen($number);
  
    // Consider every character as
    // beginning of substring
    for ($i = 0; $i < $n; $i++)
    {
        //sum of digits in 
        // current substring
        $sum = $number[$i] - '0'
  
        if ($number[$i] == '9') $count++;
  
        // One by one choose every character
        // as an ending character
        for ($j = $i + 1; $j < $n; $j++)
        {
              
            // Add current digit to sum, 
            // if sum becomes multiple of 5
            // then increment count. Let us 
            // do modular arithmetic to
            // avoid overflow for big strings
            $sum = ($sum + $number[$j] - '0') % 9;
  
            if ($sum == 0)
            $count++;
        }
    }
    return $count;
}
  
    // Driver Code
    echo count9s("4189"),"\n";
    echo count9s("1809");
      
// This code is contributed by ajit
?>



Output:

3
5

Time complexity of the above program is O(n2). Please let me know if there is a better solution.

Given a number as a string, find the number of contiguous subsequences which recursively add up to 9 | Set 2

This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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