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. 
 

Output: 
 

8
5
3

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

Auxiliary Space: O(1).