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

• Difficulty Level : Hard
• Last Updated : 29 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: 909
Output: 5
There are 5 substrings which recursively add
to nine, 9, 90, 909, 09, 9```

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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<-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 ``#include ``using` `namespace` `std;` `int` `count9s(``char` `number[])``{``    ``int` `n = ``strlen``(number);` `    ``// to store no. of previous encountered modular sums``    ``int` `d;``    ``memset``(d, 0, ``sizeof``(d));` `    ``// no. of modular sum(==0) encountered till now = 1``    ``d = 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`

## 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`

## 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``;`` ` `        ``// no. of modular sum(==0) encountered till now = 1``        ``d = 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"``));``    ``}``}`

## Javascript

 ``

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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.