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

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 <bits/stdc++.h>
using namespace std;

// Function that returns true
// if n is prime
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
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
int main()
{
    string str = "geeksforgeeks";
    int n = str.size();

    cout << sumAscii(str, n);

    return 0;
}

Java

// Java implementation of the approach
import 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 prime
def 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 positions
def 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 code
if __name__ == "__main__" :

    string = "geeksforgeeks";
    n = len(string);

    print(sumAscii(string, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using 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

<script>

// Javascript implementation of the approach

// Function that returns true
// if n is prime
function isPrime(n)
{
    if (n == 0 || n == 1)
        return false;
    for (let 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
function sumAscii(str, n)
{
    // To store the sum
    let sum = 0;

    // For every character
    for (let 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 += str.charCodeAt(i);
    }

    return sum;
}

// Driver code
    let str = "geeksforgeeks";
    let n = str.length;

    document.write(sumAscii(str, n));

</script>

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.