Non-repeating Primes

Given an array arr[] containing repetitive prime and nonprime numbers, the task is to find the prime numbers occurring only once.


Input: arr[] = {2, 3, 4, 6, 7, 9, 7, 23, 21, 2, 3}
Output: 23
In the given array, 23 is the only prime number which appears once.

Input: arr[] = {17, 19, 7, 5, 29, 5, 2, 2, 7, 17, 19}
Output: 29
In the given array, 29 is the only prime number which appears once.

Naive Approach: To solve the problem mentioned above the solution is to check every element if it is prime. If it is prime, then check is it appears only once or not. Once a prime element with a single occurrence is found, print it.

Time complexity: O(N2)

Efficient Approach: The above approach can be further optimised using Sieve and Hashing algorithm.

  1. Precompute and store prime numbers using Sieve in a Hash Table.
  2. Also create a HashMap to store the numbers with their frequency.
  3. Traverse all elements in the array one by one and:
    • Check if the current number is prime or not, using the Sieve Hash Table in O(1).
    • If the number is prime, then increase its frequency in the HashMap.
  4. Traverse the HashMap, and print all the numbers which have the frequency 1.

Below is the implementation of the above approach:





// Java program to find
// Non-repeating Primes
import java.util.*;
class GFG {
    // Function to find count of prime
    static Vector<Boolean> findPrimes(
        int arr[], int n)
        // Find maximum value in the array
        int max_val = Arrays
        // Find and store all prime numbers
        // up to max_val using Sieve
        // 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<Boolean> prime
            = new Vector<>(max_val + 1);
        for (int i = 0; i < max_val + 1; i++)
            prime.add(i, Boolean.TRUE);
        // Remaining part of SIEVE
        prime.add(0, Boolean.FALSE);
        prime.add(1, Boolean.FALSE);
        for (int p = 2; p * p <= max_val; p++) {
            // If prime[p] is not changed,
            // then it is a prime
            if (prime.get(p) == true) {
                // Update all multiples of p
                for (int i = p * 2;
                     i <= max_val;
                     i += p)
                    prime.add(i, Boolean.FALSE);
        return prime;
    // Function to print
    // Non-repeating primes
    static void nonRepeatingPrimes(
        int arr[], int n)
        // Precompute primes using Sieve
        Vector<Boolean> prime
            = findPrimes(arr, n);
        // Create HashMap to store
        // frequency of prime numbers
        HashMap<Integer, Integer> mp
            = new HashMap<>();
        // Traverse through array elements and
        // Count frequencies of all primes
        for (int i = 0; i < n; i++) {
            if (prime.get(arr[i]))
                if (mp.containsKey(arr[i]))
                           mp.get(arr[i]) + 1);
                    mp.put(arr[i], 1);
        // Traverse through map and
        // print non repeating primes
        for (Map.Entry<Integer, Integer>
                 entry : mp.entrySet()) {
            if (entry.getValue() == 1)
    // Driver code
    public static void main(String[] args)
        int arr[] = { 2, 3, 4, 6, 7, 9,
                      7, 23, 21, 3 };
        int n = arr.length;
        nonRepeatingPrimes(arr, n);




Time complexity: O(O(n*log(log(n))))
Auxiliary Space: O(K), where K is the largest value in the array.

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