# 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);` `    ``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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