# Find the sum of the ascii values of characters which are present at prime positions

• Last Updated : 28 May, 2022

Given string str of size N, the task is to find the sum of all ASCII values of the characters which are present at prime positions.
Examples:

Input: str = “abcdef”
Output: 298
‘b’, ‘c’ and ‘e’ are the only characters which are
at prime positions i.e. 2, 3 and 5 respectively.
And sum of their ASCII values is 298.
Input: str = “geeksforgeeks”
Output: 644

Approach: An efficient approach is to traverse through the whole string and find if the particular position is prime or not. If the position of the current character is prime then add the ASCII value of the character to the answer.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function that returns true// if n is primebool isPrime(int n){    if (n == 0 || n == 1)        return false;    for (int i = 2; i * i <= n; i++)        if (n % i == 0)            return false;     return true;} // Function to return the sum// of the ascii values of the characters// which are present at prime positionsint sumAscii(string str, int n){    // To store the sum    int sum = 0;     // For every character    for (int i = 0; i < n; i++) {         // If current position is prime        // then add the ASCII value of the        // character at the current position        if (isPrime(i + 1))            sum += (int)(str[i]);    }     return sum;} // Driver codeint main(){    string str = "geeksforgeeks";    int n = str.size();     cout << sumAscii(str, n);     return 0;}

## Java

 // Java implementation of the approachimport java.util.*; class GFG{     // Function that returns true    // if n is prime    static boolean isPrime(int n)    {        if (n == 0 || n == 1)        {            return false;        }        for (int i = 2; i * i <= n; i++)        {            if (n % i == 0)            {                return false;            }        }         return true;    }     // Function to return the sum    // of the ascii values of the characters    // which are present at prime positions    static int sumAscii(String str, int n)    {        // To store the sum        int sum = 0;         // For every character        for (int i = 0; i < n; i++)        {             // If current position is prime            // then add the ASCII value of the            // character at the current position            if (isPrime(i + 1))            {                sum += (int) (str.charAt(i));            }        }         return sum;    }     // Driver code    public static void main(String[] args)    {        String str = "geeksforgeeks";        int n = str.length();         System.out.println(sumAscii(str, n));    }} // This code contributed by Rajput-Ji

## Python3

 # Python3 implementation of the approach from math import sqrt # Function that returns true# if n is primedef isPrime(n) :         if (n == 0 or n == 1) :        return False;             for i in range(2, int(sqrt(n)) + 1) :        if (n % i == 0):            return False;     return True;  # Function to return the sum# of the ascii values of the characters# which are present at prime positionsdef sumAscii(string, n) :     # To store the sum    sum = 0;     # For every character    for i in range(n) :         # If current position is prime        # then add the ASCII value of the        # character at the current position        if (isPrime(i + 1)) :            sum += ord(string[i]);     return sum;  # Driver codeif __name__ == "__main__" :     string = "geeksforgeeks";    n = len(string);     print(sumAscii(string, n)); # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approachusing System; class GFG{     // Function that returns true    // if n is prime    static bool isPrime(int n)    {        if (n == 0 || n == 1)        {            return false;        }                 for (int i = 2; i * i <= n; i++)        {            if (n % i == 0)            {                return false;            }        }         return true;    }     // Function to return the sum    // of the ascii values of the characters    // which are present at prime positions    static int sumAscii(string str, int n)    {        // To store the sum        int sum = 0;         // For every character        for (int i = 0; i < n; i++)        {             // If current position is prime            // then add the ASCII value of the            // character at the current position            if (isPrime(i + 1))            {                sum += (int) (str[i]);            }        }         return sum;    }     // Driver code    public static void Main()    {        string str = "geeksforgeeks";        int n = str.Length;         Console.WriteLine(sumAscii(str, n));    }} // This code contributed by anuj_67..

## Javascript



Output:

644

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

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