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# 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 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

 ``

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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