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# Count characters in a string whose ASCII values are prime

• Last Updated : 12 Sep, 2022

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 :
‘g’, ‘e’ and ‘k’ are the only characters whose ASCII values are prime i.e. 103, 101 and 107 respectively.

Input: S = “abcdefghijklmnopqrstuvwxyz”
Output:

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 ` `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

 ``

Output

`8`

Complexity Analysis:

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

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