# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find sum of elements ` `// in an array having prime frequency ` `#include ` `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; ` `} `

## Python3

# 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

Output:

```15
```

My Personal Notes arrow_drop_up