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 remembering 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

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[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"));
    }
}

Javascript

<script>
// Javascript program to count substrings with recursive sum equal to 9

    function count9s(number)
    {
        let n = number.length;

        // to store no. of previous encountered modular sums
        let d = new Array(9);
        for(let i=0;i<d.length;i++)
        {
            d[i]=0;
        }

        // no. of modular sum(==0) encountered till now = 1
        d[0] = 1;
        let result = 0;

        let mod_sum = 0, continuous_zero = 0;
        for (let 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
    document.write(count9s("01809")+"<br>");
    document.write(count9s("1809")+"<br>");
    document.write(count9s("4189")+"<br>");


    //This code is contributed by avanitrachhadiya2155

</script>

Output:

8
5
3

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

Auxiliary Space: O(1).
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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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