Sum of elements in an array having prime frequency

Given an array arr, the task is to find the sum of the elements which have prime frequencies in the array.
Note: 1 is neither prime nor composite.

Examples:

Input: arr[] = {5, 4, 6, 5, 4, 6}
Output: 15
All the elements appear 2 times which is a prime
So, 5 + 4 + 6 = 15

Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 5
Only 2 and 3 appears prime number of times i.e. 3 and 5 respectively.
So, 2 + 3 = 5



Approach:

  • Traverse the array and store the frequencies of all the elements in a map.
  • Build Sieve of Eratosthenes which will be used to test the primality of a number in O(1) time.
  • Calculate the sum of elements having prime frequency using the Sieve array calculated in the previous step.

Below is the implementation of the above approach:

C++

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// C++ program to find sum of elements
// in an array having prime frequency
#include <bits/stdc++.h>
using namespace std;
  
// Function to create Sieve to check primes
void SieveOfEratosthenes(bool prime[], int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    for (int p = 2; p * p <= p_size; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2; i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the sum of elements
// in an array having prime frequency
int sumOfElements(int arr[], int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
  
    SieveOfEratosthenes(prime, n + 1);
  
    int i, j;
  
    // Map is used to store
    // element frequencies
    unordered_map<int, int> m;
    for (i = 0; i < n; i++)
        m[arr[i]]++;
  
    int sum = 0;
  
    // Traverse the map using iterators
    for (auto it = m.begin(); it != m.end(); it++) {
  
        // Count the number of elements
        // having prime frequencies
        if (prime[it->second]) {
            sum += (it->first);
        }
    }
  
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 5, 4, 6, 5, 4, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumOfElements(arr, n);
    return 0;
}

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Java

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// Java program to find sum of elements
// in an array having prime frequency
import java.util.*;
  
class GFG 
{
  
    // Function to create Sieve to check primes
    static void SieveOfEratosthenes(boolean prime[], int p_size)
    {
        // False here indicates
        // that it is not prime
        prime[0] = false;
        prime[1] = false;
      
        for (int p = 2; p * p <= p_size; p++)
        {
      
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p]) 
            {
      
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i <= p_size; i += p)
                    prime[i] = false;
            }
        }
    }
      
    // Function to return the sum of elements
    // in an array having prime frequency
    static int sumOfElements(int arr[], int n)
    {
        boolean prime[] = new boolean[n + 1];
        Arrays.fill(prime, true);
      
        SieveOfEratosthenes(prime, n + 1);
      
        int i, j;
      
        // Map is used to store
        // element frequencies
        HashMap<Integer, Integer> m = new HashMap<>();
        for (i = 0; i < n; i++) 
        {
            if(m.containsKey(arr[i]))
                m.put(arr[i], m.get(arr[i]) + 1);
            else
                m.put(arr[i], 1);
        }
      
        int sum = 0;
      
        // Traverse the map
        for (Map.Entry<Integer, Integer> entry : m.entrySet()) 
        {
            int key = entry.getKey();
            int value = entry.getValue();
              
            // Count the number of elements
            // having prime frequencies
            if (prime[value]) 
            {
                sum += (key);
            }
        }
      
        return sum;
    }
      
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 5, 4, 6, 5, 4, 6 };
        int n = arr.length;
      
        System.out.println(sumOfElements(arr, n));
    }
}
  
// This code is contributed by ghanshyampandey

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Python3

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# Python3 program to find Sum of elements
# in an array having prime frequency
import math as mt
  
# Function to create Sieve to
# check primes
def SieveOfEratosthenes(prime, p_size):
      
    # False here indicates
    # that it is not prime
    prime[0] = False
    prime[1] = False
  
    for p in range(2, mt.ceil(mt.sqrt(p_size + 1))):
  
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
  
            # Update all multiples of p,
            # set them to non-prime
            for i in range(p * 2, p_size + 1, p):
                prime[i] = False
          
# Function to return the Sum of elements
# in an array having prime frequency
def SumOfElements(arr, n):
    prime = [True for i in range(n + 1)]
    SieveOfEratosthenes(prime, n + 1)
  
    i, j = 0, 0
  
    # Map is used to store
    # element frequencies
    m = dict()
    for i in range(n):
        if arr[i] in m.keys():
            m[arr[i]] += 1
        else:
            m[arr[i]] = 1
              
    Sum = 0
  
    # Traverse the map using iterators
    for i in m:
          
        # Count the number of elements
        # having prime frequencies
        if (prime[m[i]]):
            Sum += (i)
      
    return Sum
  
# Driver code
arr = [5, 4, 6, 5, 4, 6 ]
n = len(arr)
print(SumOfElements(arr, n))
  
# This code is contributed
# by Mohit kumar 29

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Output:

15


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