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

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  • Last Updated : 08 Dec, 2022
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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<bits/stdc++.h>
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 <n; i++)
        // finding sum of ASCII value of first
        // substring
        d += s[i] ;
     
    if (d % k == 0)
        count += 1 ;
     
    for (i = k; i < n; i++)
    {
     
        // Using sliding window technique to
        // find sum of ASCII value of rest of
        // the substring
        int prev = s[i-k];
        d -= prev;
        d += s[i];
     
     
     
        // checking if sum is divisible by k
        if (d % k == 0)
        count += 1;
    }
     
    return count ;
    }
     
    // Driver code
    int main()
    {
         
        string s = "bcgabc" ;
        int k = 3 ;
        int ans = count(s, k);
        cout<<(ans);
    }
    // This code is contributed by
    // Sahil_Shelangia

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 <n; i++)
        // finding sum of ASCII value of first
        // substring
        d += s.charAt(i) ;
     
    if (d % k == 0)
        count += 1 ;
     
    for (i = k; i < n; i++)
    {
     
        // Using sliding window technique to
        // find sum of ASCII value of rest of
        // the substring
        int prev = s.charAt(i-k) ;
        d -= prev ;
        d += s.charAt(i) ;
     
     
     
        // checking if sum is divisible by k
        if (d % k == 0)
        count += 1 ;
    }
     
    return count ;
    }
     
    // Driver code
    public static void main(String[]args) {
         
        String s = "bcgabc" ;
        int k = 3 ;
        int ans = count(s, k);
        System.out.println(ans);
    }
    // This code is contributed by Ryuga
}

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 <n; i++)
        // finding sum of ASCII value of first
        // substring
        d += s[i] ;
      
    if (d % k == 0)
        count += 1 ;
      
    for (i = k; i < n; i++)
    {
      
        // Using sliding window technique to
        // find sum of ASCII value of rest of
        // the substring
        int prev = s[i-k] ;
        d -= prev ;
        d += s[i] ;
      
      
      
        // checking if sum is divisible by k
        if (d % k == 0)
        count += 1 ;
    }
      
    return count ;
    }
      
    // Driver code
    public static void Main() {
          
        string s = "bcgabc" ;
        int k = 3 ;
        int ans = count(s, k);
        Console.Write(ans);
    }
   
}

Javascript




<script>
 
// JavaScript program to find number of
// substrings of length k whose sum of
// ASCII value of characters is divisible by k
function count(s, k)
{
     
    // Finding length of string
    var n = s.length;
    var d = 0, i;
    var count = 0;
     
    for(i = 0; i < n; i++)
     
        // Finding sum of ASCII value of first
        // substring
        d += s[i].charCodeAt(0);
         
        if (d % k === 0)
        {
            count += 1;
        }
         
        for(i = k; i < n; i++)
        {
             
            // Using sliding window technique to
            // find sum of ASCII value of rest of
            // the substring
            var prev = s[i - k];
            d -= prev.charCodeAt(0);
            d += s[i].charCodeAt(0);
             
            // checking if sum is divisible by k
            if (d % k === 0)
                count += 1;
        }
         
    return count;
}
 
// Driver code
var s = "bcgabc";
var k = 3;
var ans = count(s, k);
 
document.write(ans);
 
// This code is contributed by rdtank
 
</script>

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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