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++ program to implement the above approach
#include <bits/stdc++.h>
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 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

# 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# 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.

<script>
// Javascript program to implement the above approach


// Function to find the required absolute difference
function xorOr(arr, N) {
    // Store the count of odds & evens at odd
    // and even indices respectively
    let evens = 0, odds = 0;

    // Traverse the array to count even/odd
    for (let 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
let arr = [3, 4, 1, 5];
let N = arr.length;
document.write(xorOr(arr, N));

// This code is contributed by gfgking.
</script>

0

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