# Number of substrings divisible by 6 in a string of integers

Datetime:2017-04-14 05:23:43         Topic: Dynamic Programming  C++          Share        Original >>

Given a string consisting of integers 0 to 9. The task is to count the number of substrings which when convert into integer are divisible by 6. Substring does not contain leading zeroes.

Examples:

```Input : s = "606".
Output : 5
Substrings "6", "0", "6", "60", "606"
are divisible by 6.

Input : s = "4806".
Output : 5
"0", "6", "48", "480", "4806" are
substring which are divisible by 6.```

Method 1: (Brute Force) The idea is to find all the substrings of the given string and check if substring is divisible by 6 or not .

Time Complexity: O(n 2 ).

Method 2:(Dynamic Programming) As discussed in Check if a large number is divisible by 6 or not . A number is divisible by 6 if last digit is divisible by 2 and sum of digits is divisible by 3.

The idea is to use Dynamic Programming, which enables us to compute answer quickly and efficiently by tracking previously computed answers and using these stored answer instead of recomputing values.

Let f(i, m) be the number of strings starting at index i and sum of their digits modulo 3 (so far) is m and number it represents is even . So, our answer would be

Let x be the i th digit in the string. From f(i, m) we need to find all the even substrings that start in i + 1.

Also, we will get an extra substring if (x + m) itself is divisible by 3 and x is even. So, we get recurrence relation

```// We initially pass m (sum modulo 3 so far) as 0
f(i, m) = ((x + m)%3 == 0 and x%2 == 0) +
f(i + 1, (m + x)%3)  // Recursive```

By memorizing the states, we get O(n) solution.

Below is C++ implementation of this approach:

```// C++ program to calculate number of substring
// divisible by 6.
#include <bits/stdc++.h>
#define MAX 100002
using namespace std;

// Return the number of substring divisible by 6
// and starting at index i in s[] and previous sum
// of digits modulo 3 is m.
int f(int i, int m, char s[], int memoize[][3])
{
// End of the string.
if (i == strlen(s))
return 0;

// If already calculated, return the
// stored value.
if (memoize[i][m] != -1)
return memoize[i][m];

// Converting into integer.
int x = s[i] - '0';

// Increment result by 1, if current digit
// is divisible by 2 and sum of digits is
// divisible by 3.
// And recur for next index with new modulo.
int ans = ((x+m)%3 == 0 && x%2 == 0) +
f(i+1, (m+x)%3, s, memoize);

return memoize[i][m] = ans;
}

// Returns substrings divisible by 6.
int countDivBy6(char s[])
{
int n = strlen(s);

// For storing the value of all states.
int memoize[n+1][3];
memset(memoize, -1, sizeof memoize);

int ans = 0;
for (int i = 0; i < strlen(s); i++)
{
// If string contain 0, increment count by 1.
if (s[i] == '0')
ans++;

// Else calculate using recursive function.
// Pass previous sum modulo 3 as 0.
else
ans += f(i, 0, s, memoize);
}

return ans;
}

// Driven Program
int main()
{
char s[] = "4806";

cout << countDivBy6(s) << endl;

return 0;
}```

Output:

Time Complexity: O(n).

This article is contributed by Anuj Chauhan . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.