Count numbers which can be represented as sum of same parity primes

Given an arr[] of positive integers you have to count how many numbers can be represented as sum of same parity prime numbers(can be same)

Examples:

Input : arr[] = {1, 3, 4, 6}
Output : 2
Explanation: 4 = 2+2, 6 = 3+3

Input : arr[] = {4, 98, 0, 36, 51}
Output : 3

1. If two numbers of same parity are added then they would be always even, so all odd numbers in the array can never contribute to answer. 
2. Talking about 0 and 2 both cannot be represented by sum of same parity prime numbers. 
3. Rest of all numbers will contribute to the answer (Refer https://www.geeksforgeeks.org/program-for-goldbachs-conjecture-two-primes-with-given-sum/)

So, we have to just iterate over the entire array and find out number of even elements not equal to 0 and 2.  

C++

#include <bits/stdc++.h>
using namespace std;

// Function to calculate count
int calculate(int* array, int size)
{
    int count = 0;

    for (int i = 0; i < size; i++)
        if (array[i] % 2 == 0 &&
            array[i] != 0 &&
            array[i] != 2)
            count++;

    return count;
}

// Driver Code
int main()
{
    int a[] = { 1, 3, 4, 6 };
    int size = sizeof(a) / sizeof(a[0]);
    cout << calculate(a, size);
}

Java

// Java program to Count numbers
// which can be represented as
// sum of same parity primes
import java.util.*;

class GFG
{
// Function to calculate count
public static int calculate(int ar[],
                            int size)
{
    int count = 0;

    for (int i = 0; i < size; i++)
        if (ar[i] % 2 == 0 &&
            ar[i] != 0 &&
            ar[i] != 2)
            count++;

    return count;
}

// Driver code
public static void main (String[] args)
{
    int a[] = { 1, 3, 4, 6 };
    int size = a.length;
    System.out.print(calculate(a, size));
}
}

// This code is contributed
// by ankita_saini

Python3

# Function to calculate count
def calculate(array, size):

    count = 0

    for i in range(size):
        if (array[i] % 2 == 0 and
            array[i] != 0 and
            array[i] != 2 ):
            count += 1

    return count

# Driver Code
if __name__ == "__main__":
    a = [ 1, 3, 4, 6 ]
    size = len(a)
    print(calculate(a, size))

# This code is contributed
# by ChitraNayal

C#

// C# program to Count numbers
// which can be represented as
// sum of same parity primes
using System;

class GFG
{
// Function to calculate count
public static int calculate(int []ar,
                            int size)
{
    int count = 0;

    for (int i = 0; i < size; i++)
        if (ar[i] % 2 == 0 &&
            ar[i] != 0 &&
            ar[i] != 2)
            count++;

    return count;
}

// Driver code
static public void Main (String []args)
{
    int []a = { 1, 3, 4, 6 };
    int size = a.Length;
    Console.WriteLine(calculate(a, size));
}
}

// This code is contributed
// by Arnab Kundu

PHP

<?php
// Function to calculate count
function calculate(&$array, $size)
{
    $count = 0;

    for ($i = 0; $i < $size; $i++)
        if ($array[$i] % 2 == 0 &&
            $array[$i] != 0 &&
            $array[$i] != 2)
            $count++;

    return $count;
}

// Driver Code
$a = array(1, 3, 4, 6 );
$size = sizeof($a);
echo calculate($a, $size);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to Count numbers
// which can be represented as
// sum of same parity primes

// Function to calculate count
function calculate(ar, size)
{
    var count = 0;

    for(i = 0; i < size; i++)
        if (ar[i] % 2 == 0 &&
            ar[i] != 0 && ar[i] != 2)
            count++;

    return count;
}

// Driver code
var a = [ 1, 3, 4, 6 ];
var size = a.length;

document.write(calculate(a, size));

// This code is contributed by todaysgaurav

</script>

Output:

2

Time complexity: O(n) where n is the size of the given array
Auxiliary space: O(1)