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 0sExplanation:
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++
#include <iostream>
#include <cstring>
using namespace std;
int count9s(char number[])
{
int n = strlen(number);
int d[9];
memset(d, 0, sizeof(d));
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'))
continuous_zero++;
else
continuous_zero=0;
mod_sum += int(number[i] - '0');
mod_sum %= 9;
result+=d[mod_sum];
d[mod_sum]++;
result -= continuous_zero;
}
return result;
}
int main()
{
cout << count9s("01809") << endl;
cout << count9s("1809") << endl;
cout << count9s("4189");
return 0;
}
|
Java
class GFG {
static int count9s(char number[]) {
int n = number.length;
int d[] = new int[9];
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)
{
continuous_zero++;
} else
{
continuous_zero = 0;
}
mod_sum += (number[i] - '0');
mod_sum %= 9;
result += d[mod_sum];
d[mod_sum]++;
result -= continuous_zero;
}
return result;
}
public static void main(String[] args) {
System.out.println(count9s("01809".toCharArray()));
System.out.println(count9s("1809".toCharArray()));
System.out.println(count9s("4189".toCharArray()));
}
}
|
Python3
def count9s(number):
n = len(number)
d = [0 for i in range(9)]
d[0] = 1
result = 0
mod_sum = 0
continuous_zero = 0
for i in range(n):
if (ord(number[i]) - ord('0') == 0):
continuous_zero += 1
else:
continuous_zero = 0
mod_sum += ord(number[i]) - ord('0')
mod_sum %= 9
result += d[mod_sum]
d[mod_sum] += 1
result -= continuous_zero
return result
if __name__ == '__main__':
print(count9s("01809"))
print(count9s("1809"))
print(count9s("4189"))
|
C#
using System;
class GFG {
static int count9s(string number) {
int n = number.Length;
int[] d = new int[9];
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)
{
continuous_zero++;
} else
{
continuous_zero = 0;
}
mod_sum += (number[i] - '0');
mod_sum %= 9;
result += d[mod_sum];
d[mod_sum]++;
result -= continuous_zero;
}
return result;
}
public static void Main() {
Console.WriteLine(count9s("01809"));
Console.WriteLine(count9s("1809"));
Console.WriteLine(count9s("4189"));
}
}
|
Javascript
<script>
function count9s(number)
{
let n = number.length;
let d = new Array(9);
for(let i=0;i<d.length;i++)
{
d[i]=0;
}
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)
{
continuous_zero++;
} else
{
continuous_zero = 0;
}
mod_sum += (number[i] - '0');
mod_sum %= 9;
result += d[mod_sum];
d[mod_sum]++;
result -= continuous_zero;
}
return result;
}
document.write(count9s("01809")+"<br>");
document.write(count9s("1809")+"<br>");
document.write(count9s("4189")+"<br>");
</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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