Comprobar si las frecuencias de todos los caracteres de una string son primos o no

Dada una string  , la tarea es comprobar si las frecuencias de todos los caracteres de la string son primos o no. Si todas las frecuencias son primos, imprima, de  lo contrario, imprima  .
Ejemplos: 

Entrada: str = «geeksforgeeks» 
Salida: No 
 

Personaje	Frecuencia
gramo	2
mi	4
k	2
s	2
F	1
o	1
r	1

Está claro que solo las frecuencias de g, k y s son primos.
Entrada: str = “aabbbcccccc” 
Salida: Sí 

Enfoque: Encuentre las frecuencias de todos los caracteres presentes en la string y guárdelas en un mapa. Luego verifique si todas las frecuencias son primos o no, si todas las frecuencias son primos entonces imprima  otra cosa  .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// function that returns true
// if n is prime else false
bool isPrime(int n)
{
    int i;
 
    // 1 is not prime
    if (n == 1)
        return false;
 
    // check if there is any factor or not
    for (i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// function that returns true if
// the frequencies of all the
// characters of s are prime
bool check_frequency(string s)
{
    // create a map to store
    // the frequencies of characters
    map<char, int> m;
 
    for (int i = 0; i < s.length(); i++)
        // update the frequency
        m[s[i]]++;
 
    // check whether all the frequencies
    // are prime or not
    for (char ch = 'a'; ch <= 'z'; ch++)
        if (m[ch] > 0 && !isPrime(m[ch]))
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    string s = "geeksforgeeks";
 
    // if all the frequencies are prime
    if (check_frequency(s))
        cout << "Yes" << endl;
 
    else
        cout << "No" << endl;
 
    return 0;
}

import java.util.*;
 
// Java implementation of above approach
class GFG
{
 
    // function that returns true
    // if n is prime else false
    static boolean isPrime(int n)
    {
        int i;
 
        // 1 is not prime
        if (n == 1)
        {
            return false;
        }
 
        // check if there is any factor or not
        for (i = 2; i <= Math.sqrt(n); i++)
        {
            if (n % i == 0)
            {
                return false;
            }
        }
 
        return true;
    }
 
    // function that returns true if
    // the frequencies of all the
    // characters of s are prime
    static boolean check_frequency(char[] s)
    {
        // create a map to store
        // the frequencies of characters
        HashMap<Character, Integer> m = new HashMap<Character, Integer>();
 
        for (int i = 0; i < s.length; i++) // update the frequency
        {
            if (m.containsKey(s[i]))
            {
                m.put(s[i], m.get(s[i]) + 1);
            }
            else
            {
                m.put(s[i], 1);
            }
        }
 
        // check whether all the frequencies
        // are prime or not
        for (char ch = 'a'; ch <= 'z'; ch++)
        {
            if (m.get(ch) != null && m.get(ch) > 0 && !isPrime(m.get(ch)))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "geeksforgeeks";
 
        // if all the frequencies are prime
        if (check_frequency(s.toCharArray()))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
// This code contributed by Rajput-Ji

# Python3 implementation of above approach
import math as mt
 
# function that returns true
# if n is prime else false
def isPrime(n):
    i = 2
 
    # 1 is not prime
    if (n == 1):
        return False
 
    # check if there is any factor or not
    for i in range(2, mt.ceil(mt.sqrt(n))):
        if (n % i == 0):
            return False
 
    return True
 
# function that returns true if the
# frequencies of all the characters
# of s are prime
def check_frequency(s):
     
    # create a map to store
    # the frequencies of characters
    m = dict()
 
    for i in range(len(s)):
         
        # update the frequency
        if s[i] in m.keys():
            m[s[i]] += 1
        else:
            m[s[i]] = 1
             
    # check whether all the frequencies
    # are prime or not
    for ch in m:
        if m[ch] > 0 and isPrime(m[ch]) == False:
            return False
 
    return True
 
# Driver code
s = "geeksforgeeks"
 
# if all the frequencies are prime
if (check_frequency(s)):
    print("Yes")
else:
    print("No")
         
# This code is contributed
# by Mohit kumar 29

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // function that returns true
    // if n is prime else false
    static bool isPrime(int n)
    {
        int i;
 
        // 1 is not prime
        if (n == 1)
        {
            return false;
        }
 
        // check if there is any factor or not
        for (i = 2; i <= Math.Sqrt(n); i++)
        {
            if (n % i == 0)
            {
                return false;
            }
        }
 
        return true;
    }
 
    // function that returns true if
    // the frequencies of all the
    // characters of s are prime
    static bool check_frequency(char[] s)
    {
        // create a map to store
        // the frequencies of characters
        Dictionary<char, int> m = new Dictionary<char, int>();
 
        for (int i = 0; i < s.Length; i++) // update the frequency
        {
            if (m.ContainsKey(s[i]))
            {
                var c = m[s[i]]+1;
                m.Remove(s[i]);
                m.Add(s[i], c);
            }
            else
            {
                m.Add(s[i], 1);
            }
        }
 
        // check whether all the frequencies
        // are prime or not
        for (char ch = 'a'; ch <= 'z'; ch++)
        {
            if (m.ContainsKey(ch) && m[ch] > 0 &&
                                    !isPrime(m[ch]))
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s = "geeksforgeeks";
 
        // if all the frequencies are prime
        if (check_frequency(s.ToCharArray()))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

<script>
 
// Javascript implementation of above approach
 
// function that returns true
// if n is prime else false
function isPrime(n)
{
    var i;
 
    // 1 is not prime
    if (n == 1)
        return false;
 
    // check if there is any factor or not
    for (i = 2; i <= Math.sqrt(n); i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// function that returns true if
// the frequencies of all the
// characters of s are prime
function check_frequency(s)
{
    // create a map to store
    // the frequencies of characters
    var m = new Map();
 
    for (var i = 0; i < s.length; i++)
        // update the frequency
        if(m.has(s[i]))
        {
            m.set(s[i],m.get(s[i])+1);
        }
        else
        {
            m.set(s[i],1);
        }
 
    // check whether all the frequencies
    // are prime or not
    for (var ch = 'a'.charCodeAt(0); ch <= 'z'.charCodeAt(0); ch++)
        if (m.get(String.fromCharCode(ch)) > 0 && !isPrime(m.get(String.fromCharCode(ch))))
            return false;
 
    return true;
}
 
// Driver code
var s = "geeksforgeeks";
 
// if all the frequencies are prime
if (check_frequency(s))
    document.write( "Yes" );
else
    document.write( "No" );
 
// This code is contributed byrutvik_56.
</script>   

No

Complejidad de tiempo: O((len(str)) ^1/2 )

Espacio auxiliar: O(len(str))