Count characters in a string whose ASCII values are prime

Given a string S. The task is to count and print the number of characters in the string whose ASCII values are prime.

Examples:

Input: S = “geeksforgeeks” 
Output : 3 
‘g’, ‘e’ and ‘k’ are the only characters whose ASCII values are prime i.e. 103, 101 and 107 respectively.

Input: S = “abcdefghijklmnopqrstuvwxyz” 
Output: 6 

Approach: The idea is to generate all primes upto max ASCII value of character of string S using Sieve of Eratosthenes. Now, Iterate the string and get the ASCII value of each character. If the ASCII value is prime then increment the count. Finally, print the count.

Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define max_val 257

// Function to find prime characters in the string
int PrimeCharacters(string s)
{

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a Boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);

    // 0 and 1 are not primes
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    int count = 0;

    // Traverse all the characters
    for (int i = 0; i < s.length(); ++i) {
        if (prime[int(s[i])])
            count++;
    }

    return count;
}

// Driver program
int main()
{
    string S = "geeksforgeeks";

    // print required answer
    cout << PrimeCharacters(S);

    return 0;
}

Java

// Java implementation of above approach
class Solution
{
static final int max_val=257;

// Function to find prime characters in the String
static int PrimeCharacters(String s)
{

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a Boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    boolean prime[]= new boolean[max_val+1];

    //initialize the value
    for(int i=0;i<=max_val;i++)
    prime[i]=true;

    // 0 and 1 are not primes
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    int count = 0;

    // Traverse all the characters
    for (int i = 0; i < s.length(); ++i) {
        if (prime[(int)(s.charAt(i))])
            count++;
    }

    return count;
}

// Driver program
public static void main(String args[])
{
    String S = "geeksforgeeks";

    // print required answer
    System.out.print( PrimeCharacters(S));

}
}
//contributed by Arnab Kundu

Python3

# Python3 implementation of above approach

from math import sqrt

max_val = 257

# Function to find prime characters in the string
def PrimeCharacters(s) :

    # USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    # THAN OR EQUAL TO max_val
    # Create a Boolean array "prime[0..n]". A
    # value in prime[i] will finally be false
    # if i is Not a prime, else true.
    prime = [True] * (max_val + 1)

    # 0 and 1 are not primes
    prime[0] = False
    prime[1] = False
    for p in range(2, int(sqrt(max_val)) + 1) :

        # If prime[p] is not changed, then
        # it is a prime
        if (prime[p] == True) :

            # Update all multiples of p
            for i in range(2*p ,max_val + 1, p) :
                prime[i] = False

    count = 0

    # Traverse all the characters
    for i in range(len(s)) :
        if (prime[ord(s[i])]) :
            count += 1

    return count

# Driver program
if __name__ == "__main__" :

    S = "geeksforgeeks";

    # print required answer
    print(PrimeCharacters(S))

# This code is contributed by Ryuga

C#

// C# implementation of above approach
using System;
class GFG{

static readonly int max_val = 257;

// Function to find prime characters in the String
static int PrimeCharacters(String s)
{
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a Boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    bool []prime = new bool[max_val + 1];

    //initialize the value
    for(int i = 0; i <= max_val; i++)
    prime[i] = true;

    // 0 and 1 are not primes
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++)
    {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
            prime[i] = false;

        }

    }

    int count = 0;

    // Traverse all the characters
    for (int i = 0; i < s.Length; ++i)
    {
        if (prime[(int)(s[i])])
        count++;

    }
    return count;

}

// Driver Code
public static void Main()
{
    String S = "geeksforgeeks";

    // print required answer
    Console.Write( PrimeCharacters(S));

}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
      // JavaScript implementation of above approach
      const max_val = 257;

      // Function to find prime characters in the String
      function PrimeCharacters(s) {
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a Boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        var prime = new Array(max_val + 1);

        //initialize the value
        for (var i = 0; i <= max_val; i++) prime[i] = true;

        // 0 and 1 are not primes
        prime[0] = false;
        prime[1] = false;
        for (var p = 2; p * p <= max_val; p++) {
          // If prime[p] is not changed, then
          // it is a prime
          if (prime[p] === true) {
            // Update all multiples of p
            for (var i = p * 2; i <= max_val; i += p) prime[i] = false;
          }
        }

        var count = 0;

        // Traverse all the characters
        for (var i = 0; i < s.length; ++i) {
          if (prime[s[i].charCodeAt(0)]) count++;
        }
        return count;
      }

      // Driver Code
      var S = "geeksforgeeks";

      // print required answer
      document.write(PrimeCharacters(S));

      // This code is contributed by rdtank.
    </script>

Output

8

Complexity Analysis:

• Time Complexity: O(max_val*log(log(max_val)))
• Auxiliary Space: O(max_val)