# Count of sub-strings that are divisible by K

Given an integer K and a numeric string str (all the characters are from the range [‘0’, ‘9’]). The task is to count the number of sub-strings of str that are divisible by K.

Examples:

Input: str = “33445”, K = 11
Output: 3
Sub-strings that are divisible by 11 are “33”, “44” and “3344”

Input: str = “334455”, K = 11
Output: 6

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

Approach: Initialize count = 0. Take all the sub-strings of str and check whether they are divisible by K or not. If yes then update count = count + 1. Print the count in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of sub-strings ` `// of str that are divisible by k ` `int` `countSubStr(string str, ``int` `len, ``int` `k) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < len; i++) { ` `        ``int` `n = 0; ` ` `  `        ``// Take all sub-strings starting from i ` `        ``for` `(``int` `j = i; j < len; j++) { ` `            ``n = n * 10 + (str[j] - ``'0'``); ` ` `  `            ``// If current sub-string is divisible by k ` `            ``if` `(n % k == 0) ` `                ``count++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the required count ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"33445"``; ` `    ``int` `len = str.length(); ` `    ``int` `k = 11; ` `    ``cout << countSubStr(str, len, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach  ` `class` `GFG ` `{ ` ` `  `    ``// Function to return the count of sub-strings  ` `    ``// of str that are divisible by k  ` `    ``static` `int` `countSubStr(String str, ``int` `len, ``int` `k)  ` `    ``{  ` `        ``int` `count = ``0``;  ` `     `  `        ``for` `(``int` `i = ``0``; i < len; i++)  ` `        ``{  ` `            ``int` `n = ``0``;  ` `     `  `            ``// Take all sub-strings starting from i  ` `            ``for` `(``int` `j = i; j < len; j++) ` `            ``{  ` `                ``n = n * ``10` `+ (str.charAt(j) - ``'0'``);  ` `     `  `                ``// If current sub-string is divisible by k  ` `                ``if` `(n % k == ``0``)  ` `                    ``count++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the required count  ` `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String []args) ` `    ``{  ` `        ``String str = ``"33445"``;  ` `        ``int` `len = str.length();  ` `        ``int` `k = ``11``;  ` `        ``System.out.println(countSubStr(str, len, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by Ryuga `

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the count of sub-strings ` `# of str that are divisible by k ` `def` `countSubStr(``str``, l, k): ` `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range``(l): ` `        ``n ``=` `0` ` `  `        ``# Take all sub-strings starting from i ` `        ``for` `j ``in` `range``(i, l, ``1``): ` `            ``n ``=` `n ``*` `10` `+` `(``ord``(``str``[j]) ``-` `ord``(``'0'``)) ` ` `  `            ``# If current sub-string is divisible by k ` `            ``if` `(n ``%` `k ``=``=` `0``): ` `                ``count ``+``=` `1` `     `  `    ``# Return the required count ` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``str` `=` `"33445"` `    ``l ``=` `len``(``str``) ` `    ``k ``=` `11` `    ``print``(countSubStr(``str``, l, k)) ` ` `  `# This code is contributed by ` `# Sanjit_Prasad `

## C#

 `// C# implementation of above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the count of sub-strings  ` `    ``// of str that are divisible by k  ` `    ``static` `int` `countSubStr(String str, ``int` `len, ``int` `k)  ` `    ``{  ` `        ``int` `count = 0;  ` `     `  `        ``for` `(``int` `i = 0; i < len; i++)  ` `        ``{  ` `            ``int` `n = 0;  ` `     `  `            ``// Take all sub-strings starting from i  ` `            ``for` `(``int` `j = i; j < len; j++) ` `            ``{  ` `                ``n = n * 10 + (str[j] - ``'0'``);  ` `     `  `                ``// If current sub-string is divisible by k  ` `                ``if` `(n % k == 0)  ` `                    ``count++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the required count  ` `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``String str = ``"33445"``;  ` `        ``int` `len = str.Length;  ` `        ``int` `k = 11;  ` `        ``Console.WriteLine(countSubStr(str, len, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by Code_Mech `

## PHP

 ` `

Output:

```3
```

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