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 = 0Input: 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 differenceint 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 Codeint 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 approachimport java.util.*;class GFG { // Function to find the required absolute differencestatic 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 differencedef 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 Codeif __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 approachusing System;using System.Collections;class GFG { // Function to find the required absolute differencestatic 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 Codepublic 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 differencefunction 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 Codelet arr = [3, 4, 1, 5];let N = arr.length;document.write(xorOr(arr, N));// This code is contributed by gfgking.</script> |
Output
0
Time Complexity: O(N)
Auxiliary Space: O(1)
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