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

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++

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#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);
}

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Java

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// 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

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

2


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Improved By : ankita_saini, andrew1234, Ita_c



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