# 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
4 = 2+2, 6 = 3+3

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

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

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 ` `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); ` `    ``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 `

## Python 3

 `# 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

 ` `

Output:

```2
```

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