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 up to the max ASCII value of the character of string S using the 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++ 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 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 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# 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 |
<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>
|
8
Complexity Analysis:
- Time Complexity: O(max_val*log(log(max_val)))
- Auxiliary Space: O(max_val)