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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 : 13 Apr, 2021

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 ``#include ``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

 ``

## Javascript

 ``

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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