Sum of elements till the smallest index such that there are no even numbers to its right
Given an array arr[] of N integers. The task is to find the sum of elements till 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 = 16Input: arr[] = {3, 4}
Output: 7
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 till 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 righmost even element while (i >= 0 && arr[i] % 2 == 1) i--; // To store the requried 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; } |
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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 righmost even element while (i >= 0 && arr[i] % 2 == 1) i--; // To store the requried 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 |
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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 righmost even element while (i >= 0 and arr[i] % 2 == 1): i -= 1; # To store the requried 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 |
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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 righmost even element while (i >= 0 && arr[i] % 2 == 1) i--; // To store the requried 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. |
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Output:
16
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