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

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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(n2).

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 ith 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 implementation of this approach:

## C++

 `// C++ program to calculate number of substring ` `// divisible by 6. ` `#include ` `#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[]) ` `{ ` `    ``// 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]; ` `    ``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; ` `} `

## Python3

 `# Python3 program to calculate number ` `# of substring  ` ` `  `# Return the number of substring divisible  ` `# by 6 and starting at index i in s[] and  ` `# previous sum of digits modulo 3 is m.  ` `def` `f(i, m, s, memoize): ` `     `  `    ``# End of the string.  ` `    ``if` `(i ``=``=` `len``(s)): ` `        ``return` `0` ` `  `    ``# If already calculated, return  ` `    ``# the stored value.  ` `    ``if` `(memoize[i][m] !``=` `-``1``):  ` `        ``return` `memoize[i][m]  ` ` `  `    ``# Converting into integer.  ` `    ``x ``=` `ord``(s[i]) ``-` `ord``(``'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.  ` `    ``ans ``=` `(((x ``+` `m) ``%` `3` `=``=` `0` `and` `x ``%` `2` `=``=` `0``) ``+` `          ``f(i ``+` `1``, (m ``+` `x) ``%` `3``, s, memoize))  ` ` `  `    ``memoize[i][m] ``=` `ans ` `    ``return` `memoize[i][m] ` ` `  `# Returns substrings divisible by 6.  ` `def` `countDivBy6(s): ` `    ``n ``=` `len``(s) ` ` `  `    ``# For storing the value of all states. ` `    ``memoize ``=` `[[``-``1``] ``*` `3` `for` `i ``in` `range``(n ``+` `1``)] ` ` `  `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``len``(s)): ` `         `  `        ``# If string contain 0, increment  ` `        ``# count by 1.  ` `        ``if` `(s[i] ``=``=` `'0'``): ` `            ``ans ``+``=` `1` ` `  `        ``# Else calculate using recursive function.  ` `        ``# Pass previous sum modulo 3 as 0.  ` `        ``else``: ` `            ``ans ``+``=` `f(i, ``0``, s, memoize) ` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``s ``=` `"4806"` ` `  `    ``print``(countDivBy6(s)) ` ` `  `# This code is contributed by PranchalK `

Output:

```5
```

Time Complexity: O(n).