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

• Difficulty Level : Easy
• Last Updated : 08 Apr, 2021

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:

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

## 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

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## Javascript

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Output:
`2`

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