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Check whether the frequencies of all the characters in a string are prime or not
  • Last Updated : 26 Feb, 2021

Given a string str  , the task is to check if the frequencies of all the characters of the string are prime or not. If all the frequencies are prime then print Yes  otherwise print No  .
Examples: 

Input: str = “geeksforgeeks” 
Output: No 
 

CharacterFrequency
g2
e4
k2
s2
f1
o1
r1

It is clear that only the frequencies of g, k and s are prime.
Input: str = “aabbbccccc” 
Output: Yes 

Approach: Find the frequencies of all the characters present in the string and store them in a map. Then check whether all the frequencies are prime or not, if all the frequency are prime then print Yes  else No  .
Below is the implementation of the above approach: 
 

C++




// 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;
}

Java




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




# 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#




// 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 */
Output: 
No

 

Time Complexity: O((len(str))1/2)

Auxiliary Space: O(len(str))

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