# Remove characters from given string whose frequencies are a Prime Number

Given string str of length N, the task is to remove all the characters from the string whose frequencies are prime.

Examples:

Input: str = “geeksforgeeks”
Output: eeforee
Explanation:
The frequencies of characters is: { g=2, e=4, k=2,  s=2, f=1, o=1, r=1}
So, g, k and s are the characters with prime frequencies so, they are removed from the string.
The resultant string is “eeforee”

Input: str = “abcdef”
Output: abcdef
Explanation:
Since all the characters are unique with frequency as 1, so no character is removed.

Naive Approach: The simplest approach to solve this problem is to find the frequencies of every distinct character present in the string and for each character, check if its frequency is Prime or not. If the frequency is found to be prime, skip that character. Otherwise, append it to the new string formed.

Time Complexity: O( N 3/2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by using the Sieve of Eratosthenes to precompute Prime Numbers. Follow the steps below to solve the problem:

• Using Sieve of Eratosthenes, generate all prime numbers up to N and store them in the array prime[].
• Initialize a Map and store the frequency of each character.
• Then, traverse the string and find out which characters have prime frequencies with the help of the map and prime[] array.
• Ignore all those characters which have prime frequencies and store the rest in a new string.
• After the above steps, print the new string.

Below is the implementation of the above approach:

 // C++ program for the above approach #include using namespace std;   // Function to perform the seive of // eratosthenes algorithm void SieveOfEratosthenes(bool* prime, int n) {     // Initialize all entries in     // prime[] as true     for (int i = 0; i <= n; i++) {         prime[i] = true;     }       // Initialize 0 and 1 as non prime     prime[0] = prime[1] = false;       // Traversing the prime array     for (int i = 2; i * i <= n; i++) {           // If i is prime         if (prime[i] == true) {               // All multiples of i must             // be marked false as they             // are non prime             for (int j = 2; i * j <= n; j++) {                 prime[i * j] = false;             }         }     } }   // Function to remove characters which // have prime frequency in the string void removePrimeFrequecies(string s) {     // Length of the string     int n = s.length();       // Create a boolean array prime     bool prime[n + 1];       // Sieve of Eratosthenes     SieveOfEratosthenes(prime, n);       // Stores the frequency of character     unordered_map m;       // Storing the frequecies     for (int i = 0; i < s.length(); i++) {         m[s[i]]++;     }       // New string that will be formed     string new_string = "";       // Removing the characters which     // have prime frequencies     for (int i = 0; i < s.length(); i++) {           // If the character has         // prime frequency then skip         if (prime[m[s[i]]])             continue;           // Else concatenate the         // character to the new string         new_string += s[i];     }       // Print the modified string     cout << new_string; }   // Driver Code int main() {     string str = "geeksforgeeks";       // Function Call     removePrimeFrequecies(str);       return 0; }

 // Java program for the above approach import java.util.*;   class GFG{   // Function to perform the seive of // eratosthenes algorithm static void SieveOfEratosthenes(boolean[] prime,                                 int n) {           // Initialize all entries in     // prime[] as true     for(int i = 0; i <= n; i++)     {         prime[i] = true;     }           // Initialize 0 and 1 as non prime     prime[0] = prime[1] = false;           // Traversing the prime array     for(int i = 2; i * i <= n; i++)     {                   // If i is prime         if (prime[i] == true)         {                           // All multiples of i must             // be marked false as they             // are non prime             for(int j = 2; i * j <= n; j++)             {                 prime[i * j] = false;             }         }     } }   // Function to remove characters which // have prime frequency in the String static void removePrimeFrequecies(char[] s) {           // Length of the String     int n = s.length;           // Create a boolean array prime     boolean []prime = new boolean[n + 1];           // Sieve of Eratosthenes     SieveOfEratosthenes(prime, n);       // Stores the frequency of character     HashMap m = new HashMap<>();           // Storing the frequecies     for(int i = 0; i < s.length; i++)     {         if (m.containsKey(s[i]))         {             m.put(s[i], m.get(s[i]) + 1);         }         else         {             m.put(s[i], 1);         }     }           // New String that will be formed     String new_String = "";           // Removing the characters which     // have prime frequencies     for(int i = 0; i < s.length; i++)     {                   // If the character has         // prime frequency then skip         if (prime[m.get(s[i])])             continue;                       // Else concatenate the         // character to the new String         new_String += s[i];     }           // Print the modified String     System.out.print(new_String); }   // Driver Code public static void main(String[] args) {     String str = "geeksforgeeks";           // Function Call     removePrimeFrequecies(str.toCharArray()); } }   // This code is contributed by aashish1995

 // C# program for the above approach using System; using System.Collections.Generic;   class GFG{   // Function to perform the seive of // eratosthenes algorithm static void SieveOfEratosthenes(bool[] prime,                                 int n) {           // Initialize all entries in     // prime[] as true     for(int i = 0; i <= n; i++)     {         prime[i] = true;     }           // Initialize 0 and 1 as non prime     prime[0] = prime[1] = false;           // Traversing the prime array     for(int i = 2; i * i <= n; i++)     {                   // If i is prime         if (prime[i] == true)         {                           // All multiples of i must             // be marked false as they             // are non prime             for(int j = 2; i * j <= n; j++)             {                 prime[i * j] = false;             }         }     } }   // Function to remove characters which // have prime frequency in the String static void removePrimeFrequecies(char[] s) {           // Length of the String     int n = s.Length;           // Create a bool array prime     bool []prime = new bool[n + 1];           // Sieve of Eratosthenes     SieveOfEratosthenes(prime, n);       // Stores the frequency of character     Dictionary m = new Dictionary();           // Storing the frequecies     for(int i = 0; i < s.Length; i++)     {         if (m.ContainsKey(s[i]))         {             m[s[i]]++;         }         else         {             m.Add(s[i], 1);         }     }           // New String that will be formed     String new_String = "";           // Removing the characters which     // have prime frequencies     for(int i = 0; i < s.Length; i++)     {                   // If the character has         // prime frequency then skip         if (prime[m[s[i]]])             continue;                       // Else concatenate the         // character to the new String         new_String += s[i];     }           // Print the modified String     Console.Write(new_String); }   // Driver Code public static void Main(String[] args) {     String str = "geeksforgeeks";           // Function Call     removePrimeFrequecies(str.ToCharArray()); } }   // This code is contributed by aashish1995

Output:
eeforee

Time Complexity: O(N*log (log N))
Auxiliary Space: O(N)

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