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

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: 909 Output: 5 There are 5 substrings which recursively add to nine, 9, 90, 909, 09, 9

This article is about an optimized solution of problem stated below article :

Given a number as a string, find the number of contiguous subsequences which recursively add up to 9 | Set 1.

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.

Algorithm:

Initialize an array d of size 10 with 0 d[0]<-1 Initialize mod_sum = 0, continuous_zero = 0 for every character if character == '0'; continuous_zero++ else continuous_zero=0 compute mod_sum update result += d[mod_sum] update d[mod_sum]++ subtract those cases from result which have only 0s

Explanation:

If sum of digits from index i to j add up to 9, then sum(0 to i-1) = sum(0 to j) (mod 9).

We just have to remove cases which contain only zeroes.We can do this by remembring the no. of continuous zeroes upto this character(no. of these cases ending on this index) and subtracting them from the result.

Following is a simple implementation based on this approach.

The implementation assumes that there are can be leading 0’s in the 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` `n = ` `strlen` `(number); ` ` ` ` ` `// to store no. of previous encountered modular sums ` ` ` `int` `d[9]; ` ` ` `memset` `(d, 0, ` `sizeof` `(d)); ` ` ` ` ` `// no. of modular sum(==0) encountered till now = 1 ` ` ` `d[0] = 1; ` ` ` `int` `result = 0; ` ` ` ` ` `int` `mod_sum = 0, continuous_zero = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `if` `(!` `int` `(number[i] - ` `'0'` `)) ` `// if number is 0 increase ` ` ` `continuous_zero++; ` `// no. of continuous_zero by 1 ` ` ` `else` `// else continuous_zero is 0 ` ` ` `continuous_zero=0; ` ` ` `mod_sum += ` `int` `(number[i] - ` `'0'` `); ` ` ` `mod_sum %= 9; ` ` ` `result+=d[mod_sum]; ` ` ` `d[mod_sum]++; ` `// increase d value of this mod_sum ` ` ` `// subtract no. of cases where there ` ` ` `// are only zeroes in substring ` ` ` `result -= continuous_zero; ` ` ` `} ` ` ` `return` `result; ` `} ` ` ` `// driver program to test above function ` `int` `main() ` `{ ` ` ` `cout << count9s(` `"01809"` `) << endl; ` ` ` `cout << count9s(` `"1809"` `) << endl; ` ` ` `cout << count9s(` `"4189"` `); ` ` ` `return` `0; ` `} ` `// This code is contributed by Gulab Arora ` |

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## Java

`// Java program to count substrings with recursive sum equal to 9 ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `count9s(` `char` `number[]) { ` ` ` `int` `n = number.length; ` ` ` ` ` `// to store no. of previous encountered modular sums ` ` ` `int` `d[] = ` `new` `int` `[` `9` `]; ` ` ` ` ` `// no. of modular sum(==0) encountered till now = 1 ` ` ` `d[` `0` `] = ` `1` `; ` ` ` `int` `result = ` `0` `; ` ` ` ` ` `int` `mod_sum = ` `0` `, continuous_zero = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` `if` `((number[i] - ` `'0'` `) == ` `0` `) ` `// if number is 0 increase ` ` ` `{ ` ` ` `continuous_zero++; ` `// no. of continuous_zero by 1 ` ` ` `} ` `else` `// else continuous_zero is 0 ` ` ` `{ ` ` ` `continuous_zero = ` `0` `; ` ` ` `} ` ` ` `mod_sum += (number[i] - ` `'0'` `); ` ` ` `mod_sum %= ` `9` `; ` ` ` `result += d[mod_sum]; ` ` ` `d[mod_sum]++; ` `// increase d value of this mod_sum ` ` ` `// subtract no. of cases where there ` ` ` `// are only zeroes in substring ` ` ` `result -= continuous_zero; ` ` ` `} ` ` ` `return` `result; ` ` ` `} ` ` ` `// driver program to test above function ` ` ` `public` `static` `void` `main(String[] args) { ` ` ` `System.out.println(count9s(` `"01809"` `.toCharArray())); ` ` ` `System.out.println(count9s(` `"1809"` `.toCharArray())); ` ` ` `System.out.println(count9s(` `"4189"` `.toCharArray())); ` ` ` `} ` `} ` `// This code is contributed by 29AjayKumar ` |

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## Python3

# Python 3 program to count substrings with

# recursive sum equal to 9

def count9s(number):

n = len(number)

# to store no. of previous encountered

# modular sums

d = [0 for i in range(9)]

# no. of modular sum(==0) encountered

# till now = 1

d[0] = 1

result = 0

mod_sum = 0

continuous_zero = 0

for i in range(n):

# if number is 0 increase

if (ord(number[i]) – ord(‘0’) == 0):

continuous_zero += 1 # no. of continuous_zero by 1

else:

continuous_zero = 0 # else continuous_zero is 0

mod_sum += ord(number[i]) – ord(‘0’)

mod_sum %= 9

result += d[mod_sum]

d[mod_sum] += 1 # increase d value of this mod_sum

# subtract no. of cases where there

# are only zeroes in substring

result -= continuous_zero

return result

# Driver Code

if __name__ == ‘__main__’:

print(count9s(“01809”))

print(count9s(“1809”))

print(count9s(“4189”))

# This code is contributed by

# Sahil_Shelangia

## C#

`// C# program to count substrings with recursive sum equal to 9 ` ` ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `count9s(` `string` `number) { ` ` ` `int` `n = number.Length; ` ` ` ` ` `// to store no. of previous encountered modular sums ` ` ` `int` `[] d = ` `new` `int` `[9]; ` ` ` ` ` `// no. of modular sum(==0) encountered till now = 1 ` ` ` `d[0] = 1; ` ` ` `int` `result = 0; ` ` ` ` ` `int` `mod_sum = 0, continuous_zero = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `if` `((number[i] - ` `'0'` `) == 0) ` `// if number is 0 increase ` ` ` `{ ` ` ` `continuous_zero++; ` `// no. of continuous_zero by 1 ` ` ` `} ` `else` `// else continuous_zero is 0 ` ` ` `{ ` ` ` `continuous_zero = 0; ` ` ` `} ` ` ` `mod_sum += (number[i] - ` `'0'` `); ` ` ` `mod_sum %= 9; ` ` ` `result += d[mod_sum]; ` ` ` `d[mod_sum]++; ` `// increase d value of this mod_sum ` ` ` `// subtract no. of cases where there ` ` ` `// are only zeroes in substring ` ` ` `result -= continuous_zero; ` ` ` `} ` ` ` `return` `result; ` ` ` `} ` ` ` `// driver program to test above function ` ` ` `public` `static` `void` `Main() { ` ` ` `Console.WriteLine(count9s(` `"01809"` `)); ` ` ` `Console.WriteLine(count9s(` `"1809"` `)); ` ` ` `Console.WriteLine(count9s(` `"4189"` `)); ` ` ` `} ` `} ` |

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**Output:**

8 5 3

Time Complexity of the above program is O(n). Program also supports leading zeroes.

This article is contributed by **Gulab Arora**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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