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

• Last Updated : 06 Jul, 2021

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 ``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

 ``
Output:
`16`

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