Recuento de formas de hacer que Array sume incluso eliminando solo un elemento

Dada una array arr[] enteros positivos, la tarea es encontrar el número de formas de convertir la suma de la array incluso si se nos permite eliminar solo un elemento.
Ejemplos:

Entrada: arr[] = { 1, 3, 3, 2 } 
Salida: 3 
Explicación: 
1. Quite 1, luego la suma es 3 + 3 + 2 = 8. 
2. Quite 3, luego la suma es 1 + 3 + 2 = 6. 
3. Quitar 3, entonces la suma es 1 + 3 + 2 = 6.
Entrada: arr[] = { 4, 8, 3, 3, 6 } 
Salida: 3 
Explicación: 
1. Quitar 4, entonces la suma es 8 + 3 + 3 + 6 = 20. 
2. Quite 8, entonces la suma es 4 + 3 + 3 + 6 = 16. 
3. Quite 6, entonces la suma es 4 + 8 + 3 + 3 = 18.

Enfoque: La observación clave de la declaración del problema anterior es:

1. Si tenemos un número impar de elementos impares, entonces la suma siempre es impar, entonces tenemos que eliminar un número impar de la array arr[] para que la suma sea par. Como tenemos que eliminar un elemento, el número total de formas de hacer que la suma sea par es el número de elementos impares en el arreglo arr[] .
2. Si tenemos un número par de elementos impares, la suma siempre es par. Dado que tenemos que eliminar un elemento para que la suma sea par, el número total de formas de hacer que la suma sea par es el número de elementos pares en el arreglo arr[]

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to find a number of ways
// to make array element sum even by
// removing one element
int find_num_of_ways(int arr[], int N)
{
    int count_even = 0, count_odd = 0;
 
    // Finding the count of even
    // and odd elements
    for (int i = 0; i < N; i++) {
        if (arr[i] % 2) {
            count_odd++;
        }
        else {
            count_even++;
        }
    }
 
    // If count_odd is odd then
    // no. of ways is count_odd
    if (count_odd % 2) {
        return count_odd;
    }
 
    // Else no. of ways is count_even
    else {
        return count_even;
    }
}
 
// Driver Code
int main()
{
 
    // Given array arr[]
    int arr[] = { 1, 3, 3, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << find_num_of_ways(arr, N);
    return 0;
}

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find a number of ways
// to make array element sum even by
// removing one element
static int find_num_of_ways(int arr[], int N)
{
    int count_even = 0, count_odd = 0;
 
    // Finding the count of even
    // and odd elements
    for(int i = 0; i < N; i++)
    {
        if (arr[i] % 2 == 1)
        {
            count_odd++;
        }
        else
        {
            count_even++;
        }
    }
 
    // If count_odd is odd then
    // no. of ways is count_odd
    if (count_odd % 2 == 1)
    {
        return count_odd;
    }
 
    // Else no. of ways is count_even
    else
    {
        return count_even;
    }
}
 
// Driver Code
public static void main (String[] args)
{
 
    // Given array arr[]
    int arr[] = { 1, 3, 3, 2 };
    int N = 4;
 
    // Function call
    System.out.print(find_num_of_ways(arr, N));
}
}
 
// This code is contributed by Ritik Bansal

# Python3 program for the above approach
 
# Function to find a number of ways
# to make array element sum even by
# removing one element
def find_num_of_ways(arr, N):
 
    count_even = 0
    count_odd = 0
 
    # Finding the count of even
    # and odd elements
    for i in range(N):
        if (arr[i] % 2):
            count_odd += 1
        else:
            count_even += 1
 
    # If count_odd is odd then
    # no. of ways is count_odd
    if (count_odd % 2):
        return count_odd
 
    # Else no. of ways is count_even
    else:
        return count_even
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 1, 3, 3, 2 ]
    N = len(arr)
 
    # Function call
    print(find_num_of_ways(arr, N))
 
# This code is contributed by mohit kumar 29

// C# program for the above approach
using System;
class GFG{
      
// Function to find a number of ways
// to make array element sum even by
// removing one element
static int find_num_of_ways(int []arr, int N)
{
    int count_even = 0, count_odd = 0;
  
    // Finding the count of even
    // and odd elements
    for(int i = 0; i < N; i++)
    {
        if (arr[i] % 2 == 1)
        {
            count_odd++;
        }
        else
        {
            count_even++;
        }
    }
  
    // If count_odd is odd then
    // no. of ways is count_odd
    if (count_odd % 2 == 1)
    {
        return count_odd;
    }
  
    // Else no. of ways is count_even
    else
    {
        return count_even;
    }
}
  
// Driver Code
public static void Main(string[] args)
{
  
    // Given array arr[]
    int []arr = { 1, 3, 3, 2 };
    int N = 4;
  
    // Function call
    Console.Write(find_num_of_ways(arr, N));
}
}
 
// This code is contributed by Rutvik

<script>
// javascript program for the above approach
 
// Function to find a number of ways
// to make array element sum even by
// removing one element
function find_num_of_ways(arr , N)
{
    var count_even = 0, count_odd = 0;
 
    // Finding the count of even
    // and odd elements
    for(i = 0; i < N; i++)
    {
        if (arr[i] % 2 == 1)
        {
            count_odd++;
        }
        else
        {
            count_even++;
        }
    }
 
    // If count_odd is odd then
    // no. of ways is count_odd
    if (count_odd % 2 == 1)
    {
        return count_odd;
    }
 
    // Else no. of ways is count_even
    else
    {
        return count_even;
    }
}
 
// Driver Code
 
// Given array arr
var arr = [ 1, 3, 3, 2 ];
var N = 4;
 
// Function call
document.write(find_num_of_ways(arr, N));
 
// This code is contributed by Amit Katiyar
</script>

3

Complejidad temporal: O(N)  
Espacio auxiliar: O(1)