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Sum of elements till the smallest index such that there are no even numbers to its right

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Given an array arr[] of N integers. The task is to find the sum of elements to the smallest index such that there are no even elements to the right of the index. Note that the array will have at least one even element.
Examples: 
 

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

 

Approach: Find the index of the rightmost even element from the array and return the sum of all the elements starting from index 0 to the found index.
Below is the implementation of the above approach: 
 

C++




// 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




// 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




# 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#




// 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.


Javascript




<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>


Output: 

16

 

Time Complexity: O(N)

Auxiliary Space: O(1)


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Last Updated : 06 Nov, 2021
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