Given a string 
, 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 
otherwise print 
.
Examples:
Input: str = “neveropen”
Output: No
Character Frequency g 2 e 4 k 2 s 2 f 1 o 1 r 1 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 else .
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 falsebool 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 primebool 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 codeint main(){ string s = "neveropen"; // 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 approachclass 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 = "neveropen"; // 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 approachimport math as mt# function that returns true# if n is prime else falsedef 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 primedef 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 codes = "neveropen"# if all the frequencies are primeif (check_frequency(s)): print("Yes")else: print("No") # This code is contributed # by Mohit kumar 29 |
C#
// C# implementation of above approachusing 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 = "neveropen"; // if all the frequencies are prime if (check_frequency(s.ToCharArray())) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript implementation of above approach// function that returns true// if n is prime else falsefunction 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 primefunction 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 codevar s = "neveropen";// if all the frequencies are primeif (check_frequency(s)) document.write( "Yes" );else document.write( "No" );// This code is contributed byrutvik_56.</script> |
Output
No
Complexity Analysis:
- Time Complexity: O((len(str))1/2)
- Auxiliary Space: O(len(str))
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