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

Last Updated : 13 Dec, 2021

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 <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

// 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# 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

<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>

Output

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

Difference between sum of K maximum even and odd array elements

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

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