Suma de elementos hasta el índice más pequeño tal que no haya números pares a su derecha

Dada una array arr[] de N enteros. La tarea es encontrar la suma de elementos al índice más pequeño tal que no haya elementos pares a la derecha del índice. Tenga en cuenta que la array tendrá al menos un elemento par.
Ejemplos: 
 

Entrada: arr[] = {2, 3, 5, 6, 3, 3} 
Salida: 16 
2 + 3 + 5 + 6 = 16
Entrada: arr[] = {3, 4} 
Salida: 7 
3 + 4 = 7 
 

Enfoque: encuentre el índice del elemento par más a la derecha de la array y devuelva la suma de todos los elementos a partir del índice 0 al índice encontrado.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the required sum
int smallestIndexsum(int arr[], int n)
{
 
    // Starting from the last index
    int i = n - 1;
 
    // Skip all odd elements and find the
    // index of the rightmost even element
    while (i >= 0 && arr[i] % 2 == 1)
        i--;
 
    // To store the required sum
    int sum = 0;
    for (int j = 0; j <= i; j++)
        sum += arr[j];
 
    return sum;
}
 
// Driver code
int main()
{
 
    int arr[] = { 2, 3, 5, 6, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << smallestIndexsum(arr, n);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
 
// Function to return the required sum
static int smallestIndexsum(int arr[], int n)
{
 
    // Starting from the last index
    int i = n - 1;
 
    // Skip all odd elements and find the
    // index of the rightmost even element
    while (i >= 0 && arr[i] % 2 == 1)
        i--;
 
    // To store the required sum
    int sum = 0;
    for (int j = 0; j <= i; j++)
        sum += arr[j];
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 5, 6, 3, 3 };
    int n = arr.length;
 
    System.out.println(smallestIndexsum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

# Python3 implementation of the approach
 
# Function to return the required sum
def smallestIndexsum(arr, n):
 
    # Starting from the last index
    i = n - 1;
 
    # Skip all odd elements and find the
    # index of the rightmost even element
    while (i >= 0 and arr[i] % 2 == 1):
        i -= 1;
 
    # To store the required sum
    sum = 0;
    for j in range(0, i + 1):
        sum += arr[j];
 
    return sum;
 
# Driver code
if __name__ == '__main__':
 
    arr = [ 2, 3, 5, 6, 3, 3 ];
    n = len(arr);
 
    print(smallestIndexsum(arr, n));
 
# This code is contributed by PrinciRaj1992

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the required sum
static int smallestIndexsum(int []arr, int n)
{
 
    // Starting from the last index
    int i = n - 1;
 
    // Skip all odd elements and find the
    // index of the rightmost even element
    while (i >= 0 && arr[i] % 2 == 1)
        i--;
 
    // To store the required sum
    int sum = 0;
    for (int j = 0; j <= i; j++)
        sum += arr[j];
 
    return sum;
}
 
// Driver code
static public void Main ()
{
    int []arr = { 2, 3, 5, 6, 3, 3 };
    int n = arr.Length;
     
    Console.Write(smallestIndexsum(arr, n));
}
}
 
// This code is contributed by ajit.

<script>
 
    // Javascript implementation of the approach
     
    // Function to return the required sum
    function smallestIndexsum(arr, n)
    {
 
        // Starting from the last index
        let i = n - 1;
 
        // Skip all odd elements and find the
        // index of the rightmost even element
        while (i >= 0 && arr[i] % 2 == 1)
            i--;
 
        // To store the required sum
        let sum = 0;
        for (let j = 0; j <= i; j++)
            sum += arr[j];
 
        return sum;
    }
     
    let arr = [ 2, 3, 5, 6, 3, 3 ];
    let n = arr.length;
       
    document.write(smallestIndexsum(arr, n));
 
</script>

16

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)