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

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++ 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;
}

Output:

3
5

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

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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