Absolute difference between sum of even elements at even indices & odd elements at odd indices in given Array

Given an array arr[] containing N elements, the task is to find the absolute difference between the sum of even elements at even indices & the count of odd elements at odd indices. Consider 1-based indexing

Examples:

Input: arr[] = {3, 4, 1, 5}
Output: 0
Explanation: Sum of even elements at even indices: 4 {4}
Sum of odd elements at odd indices: 4 {3, 1}
Absolute Difference = 4-4 = 0

Input: arr[] = {4, 2, 1, 3}
Output: 1

Approach: The task can be solved by traversing the array from left to right, keeping track of sum of odd and even elements at odd & even indices respectively. Follow the steps below to solve the problem:

• Traverse the array from left to right.
• If the current index is even, check whether the element at that index is even or not, If it’s even, add it to the sum evens.
• If the current index is odd, check whether the element at that index is odd or not, If it’s odd, add it to the sum odds.
• Return the absolute difference between odds and evens.

Below is the implementation of the above approach.

C++

 // C++ program to implement the above approach #include using namespace std;   // Function to find the required absolute difference int xorOr(int arr[], int N) {     // Store the count of odds & evens at odd     // and even indices respectively     int evens = 0, odds = 0;       // Traverse the array to count even/odd     for (int i = 0; i < N; i++) {         if ((i + 1) % 2 == 0             && arr[i] % 2 == 0)             evens += arr[i];         else if ((i + 1) % 2 != 0                  && arr[i] % 2 != 0)             odds += arr[i];     }       return abs(odds - evens); }   // Driver Code int main() {     int arr[] = { 3, 4, 1, 5 };     int N = sizeof(arr) / sizeof(arr[0]);     cout << xorOr(arr, N);     return 0; }

Java

 // Java code for the above approach import java.util.*; class GFG {     // Function to find the required absolute difference static int xorOr(int arr[], int N) {     // Store the count of odds & evens at odd     // and even indices respectively     int evens = 0, odds = 0;       // Traverse the array to count even/odd     for (int i = 0; i < N; i++) {         if ((i + 1) % 2 == 0             && arr[i] % 2 == 0)             evens += arr[i];         else if ((i + 1) % 2 != 0                  && arr[i] % 2 != 0)             odds += arr[i];     }       return Math.abs(odds - evens); }   // Driver Code     public static void main (String[] args) {        int arr[] = { 3, 4, 1, 5 };     int N = arr.length;                System.out.println(xorOr(arr, N));     } }   // This code is contributed by Potta Lokesh

Python3

 # Python code for the above approach   # Function to find the required absolute difference def xorOr(arr, N):         # Store the count of odds & evens at odd     # and even indices respectively     evens = 0;     odds = 0;       # Traverse the array to count even/odd     for i in range(N):         if ((i + 1) % 2 == 0 and arr[i] % 2 == 0):             evens += arr[i];         elif ((i + 1) % 2 != 0 and arr[i] % 2 != 0):             odds += arr[i];       return abs(odds - evens);   # Driver Code if __name__ == '__main__':     arr = [3, 4, 1, 5];     N = len(arr);       print(xorOr(arr, N));   # This code is contributed by 29AjayKumar

C#

 // C# code for the above approach using System; using System.Collections; class GFG {     // Function to find the required absolute difference static int xorOr(int []arr, int N) {         // Store the count of odds & evens at odd     // and even indices respectively     int evens = 0, odds = 0;       // Traverse the array to count even/odd     for (int i = 0; i < N; i++) {         if ((i + 1) % 2 == 0             && arr[i] % 2 == 0)             evens += arr[i];         else if ((i + 1) % 2 != 0                  && arr[i] % 2 != 0)             odds += arr[i];     }       return Math.Abs(odds - evens); }   // Driver Code public static void Main () {     int []arr = { 3, 4, 1, 5 };     int N = arr.Length;                Console.Write(xorOr(arr, N)); } }   // This code is contributed by Samim Hossain Mondal.

Javascript



Output

0

Time Complexity: O(N)
Auxiliary Space: O(1)

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