# Find number of substrings of length k whose sum of ASCII value of characters is divisible by k

• Last Updated : 08 Dec, 2022

Given a string and a number k, the task is to find the number of substrings of length k whose sum of ASCII value of characters is divisible by k.

Examples:

Input : str = “bcgabc”, k = 3
Output :
Substring “bcg” has sum of ASCII values 300 and “abc” has sum of ASCII values 294 which are divisible by 3.

Input : str = “adkf”, k = 3
Output :

Approach: First, we find the sum of ASCII value of characters of first substring of length k, then using sliding window technique subtract ASCII value of first character of the previous substring and add ASCII value of the current character. We will increase the counter at each step if sum is divisible by k.

Below is the implementation of the above approach:

## C++

 `// C++ program to find number of substrings``// of length k whose sum of ASCII value of``// characters is divisible by k``#include``using` `namespace` `std;` `int` `count(string s, ``int` `k)``{``    ` `    ``// Finding length of string``    ``int` `n = s.length();``    ``int` `d = 0 ,i;``    ``int` `count = 0 ;``    ` `    ``for` `(i = 0; i

## Java

 `// Java program to find number of substrings``// of length k whose sum of ASCII value of``// characters is divisible by k`  `public` `class` `GFG{` `    ``static` `int` `count(String s, ``int` `k){``    ` `    ``// Finding length of string``    ``int` `n = s.length() ;``    ``int` `d = ``0` `,i;``    ``int` `count = ``0` `;``    ` `    ``for` `(i = ``0``; i

## Python3

 `# Python3 program to find number of substrings``# of length k whose sum of ASCII value of``# characters is divisible by k` `def` `count(s, k):``    ` `    ``# Finding length of string``    ``n ``=` `len``(s)``    ``d, count ``=` `0``, ``0``    ``for` `i ``in` `range``(k):``        ` `        ``# finding sum of ASCII value of first``        ``# substring``        ``d ``+``=` `ord``(s[i])``        ``if` `(d ``%` `k ``=``=` `0``):``            ``count ``+``=` `1``            ``for` `i ``in` `range``(k, n):``                ` `                ``# Using sliding window technique to``                ``# find sum of ASCII value of rest of``                ``# the substring``                ``prev ``=` `ord``(s[i``-``k])``                ``d ``-``=` `prev``                ``d ``+``=` `ord``(s[i])``                ` `                ``# checking if sum is divisible by k``                ``if` `(d ``%` `k ``=``=` `0``):``                    ``count ``+``=` `1``                    ``return` `count``# Driver code``s ``=` `"bcgabc"``k ``=` `3``ans ``=` `count(s, k)``print``(ans)`

## C#

 `// C# program to find number of substrings``// of length k whose sum of ASCII value of``// characters is divisible by k`` ` ` ` `using` `System;``public` `class` `GFG{`` ` `    ``static` `int` `count(``string` `s, ``int` `k){``     ` `    ``// Finding length of string``    ``int` `n = s.Length ;``    ``int` `d = 0 ,i;``    ``int` `count = 0 ;``     ` `    ``for` `(i = 0; i

## Javascript

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Output:

`2`

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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