Máximos elementos posibles que son divisibles por 2

Dada una array de enteros arr de tamaño N . La tarea es encontrar el máximo de elementos posibles en la array que sean divisibles por 2 después de modificar la array. Uno puede realizar la siguiente operación un número arbitrario de veces (posiblemente cero veces).
 

Reemplace dos elementos cualquiera en la array con su suma.

Ejemplos: 
 

Entrada: arr = [1, 2, 3, 1, 3] 
Salida: 3 
Después de agregar elementos en los índices 0 y 2, y en los índices 3 y 4, la array se convierte en arr=[4, 2, 4].
Entrada: arr = [1, 2, 3, 4, 5] 
Salida: 3 
Después de sumar 1 y 3, la array se convierte en arr=[4, 2, 4, 5]. 

Enfoque : 
Primero, la observación es que no necesitamos modificar elementos que son divisibles por 2 (es decir, números pares). Luego nos fuimos con los números impares. La suma de dos números dará un número par que es divisible por 2.
Entonces, finalmente, el resultado será: 
 

cuenta_par + cuenta_impar /2.

A continuación se muestra la implementación del enfoque anterior: 
 

// CPP program to find maximum possible
// elements which divisible by 2
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum possible
// elements which divisible by 2
int Divisible(int arr[], int n)
{
    // To store count of even numbers
    int count_even = 0;
 
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 0)
            count_even++;
 
    // All even numbers and half of odd numbers
    return count_even + (n - count_even) / 2;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << Divisible(arr, n);
 
    return 0;
}

// Java program to find maximum possible
// elements which divisible by 2
class GFG
{
 
    // Function to find maximum possible
    // elements which divisible by 2
    static int Divisible(int arr[], int n)
    {
        // To store count of even numbers
        int count_even = 0;
     
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 == 0)
                count_even++;
     
        // All even numbers and half of odd numbers
        return count_even + (n - count_even) / 2;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5 };
     
        int n = arr.length;
     
        // Function call
        System.out.println(Divisible(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

# Python3 program to find maximum possible
# elements which divisible by 2
 
# Function to find maximum possible
# elements which divisible by 2
def Divisible(arr, n):
    # To store count of even numbers
    count_even = 0
 
    for i in range(n):
        if (arr[i] % 2 == 0):
            count_even+=1
 
    # All even numbers and half of odd numbers
    return count_even + (n - count_even) // 2
 
# Driver code
 
arr=[1, 2, 3, 4, 5]
 
n = len(arr)
 
# Function call
print(Divisible(arr, n))
 
# This code is contributed by mohit kumar 29

// C# program to find maximum possible
// elements which divisible by 2
using System;
 
class GFG
{
     
    // Function to find maximum possible
    // elements which divisible by 2
    static int Divisible(int []arr, int n)
    {
        // To store count of even numbers
        int count_even = 0;
     
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 == 0)
                count_even++;
     
        // All even numbers and half of odd numbers
        return count_even + (n - count_even) / 2;
    }
     
    // Driver code
    static public void Main ()
    {
         
        int []arr = { 1, 2, 3, 4, 5 };
        int n = arr.Length;
         
        // Function call
        Console.Write(Divisible(arr, n));
    }
}
 
// This code is contributed by ajit.

<script>
// Javascript program to find maximum possible
// elements which divisible by 2
 
// Function to find maximum possible
// elements which divisible by 2
function Divisible(arr, n)
{
    // To store count of even numbers
    let count_even = 0;
 
    for (let i = 0; i < n; i++)
        if (arr[i] % 2 == 0)
            count_even++;
 
    // All even numbers and half of odd numbers
    return count_even + parseInt((n - count_even) / 2);
}
 
// Driver code
    let arr = [ 1, 2, 3, 4, 5 ];
 
    let n = arr.length;
 
    // Function call
    document.write(Divisible(arr, n));
 
</script>

3

Complejidad temporal: O(N).
Espacio Auxiliar : O(1).