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 : 2
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 : 1
Brute Force Approach:
The brute force approach to solve this problem is to generate all possible substrings of length k from the given string and calculate the sum of ASCII values of characters for each substring. Then check whether the sum is divisible by k or not. If it is divisible by k, then count it as a valid substring. Finally, return the count of valid substrings.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>using namespace std;int count(string s, int k){ int n = s.length(); int cnt = 0; for (int i = 0; i <= n - k; i++) { int sum = 0; for (int j = i; j < i + k; j++) sum += s[j]; if (sum % k == 0) cnt++; } return cnt;}int main(){ string s = "adkf"; int k = 3; int ans = count(s, k); cout << ans; return 0;} |
Java
import java.util.*;public class Main { public static int count(String s, int k) { int n = s.length(); int cnt = 0; for (int i = 0; i <= n - k; i++) { int sum = 0; for (int j = i; j < i + k; j++) sum += s.charAt(j); if (sum % k == 0) cnt++; } return cnt; } public static void main(String[] args) { String s = "adkf"; int k = 3; int ans = count(s, k); System.out.println(ans); }} |
Python3
def count(s, k): n = len(s) cnt = 0 for i in range(n - k + 1): _sum = 0 for j in range(i, i + k): _sum += ord(s[j]) if _sum % k == 0: cnt += 1 return cnts = "adkf"k = 3ans = count(s, k)print(ans) |
C#
using System;class GFG { public static int Count(string s, int k) { int n = s.Length; int cnt = 0; for (int i = 0; i <= n - k; i++) { int sum = 0; for (int j = i; j < i + k; j++) { sum += s[j]; } if (sum % k == 0) { cnt++; } } return cnt; } public static void Main(string[] args) { string s = "adkf"; int k = 3; int ans = Count(s, k); Console.WriteLine(ans); }} |
Javascript
// JavaScript codefunction count(s, k) { let n = s.length; let cnt = 0; for (let i = 0; i <= n - k; i++) { let sum = 0; for (let j = i; j < i + k; j++) { sum += s.charCodeAt(j); } if (sum % k === 0) { cnt++; } } return cnt;}let s = "adkf";let k = 3;let ans = count(s, k);console.log(ans); |
Output
1
Time Complexity: O(N^2)
Space Complexity: O(1)
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 codes = "bcgabc"k = 3ans = 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 kfunction 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 codevar 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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