Given an array arr of size N. The task is to find the XOR of the first half (N/2) elements and second half elements (N – N/2) of an array.
Examples:
Input: arr[]={20, 30, 50, 10, 55, 15, 42}
Output: 56, 24
Explanation:
XOR of first half elements 20 ^ 30 ^ 50 is 56
Xor of second half elements 10 ^ 55 ^ 15 ^ 42 is 24
Input: arr[]={50, 45, 36, 6, 8, 87}
Output: 59, 89
Approach:
- Initialize FirstHalfXOR and SecondHalfXOR as 0.
- Traverse the array and XOR elements in FirstHalfXOR until the current index is less than N/2.
- Otherwise, XOR elements in SecondHalfXOR.
Below is the implementation of the above approach:
C++
// C++ program to find the xor of // the first half elements and // second half elements of an array #include <iostream>using namespace std;// Function to find the xor of // the first half elements and // second half elements of an array void XOROfElements(int arr[], int n){ int FirstHalfXOR = 0; int SecondHalfXOR = 0; for(int i=0; i < n; i++){ // xor of elements in FirstHalfXOR if (i < n / 2) FirstHalfXOR ^= arr[i]; // xor of elements in SecondHalfXOR else SecondHalfXOR ^= arr[i]; } cout << FirstHalfXOR << "," << SecondHalfXOR << endl; }// Driver Code int main() { int arr[] = {20, 30, 50, 10, 55, 15, 42}; int N = sizeof(arr)/sizeof(arr[0]); // Function call XOROfElements(arr, N); return 0;}// This code is contributed by AnkitRai01 |
Java
// Java program to find the xor of // the first half elements and // second half elements of an array class GFG{// Function to find the xor of // the first half elements and // second half elements of an array static void XOROfElements(int arr[], int n){ int FirstHalfXOR = 0; int SecondHalfXOR = 0; for(int i = 0; i < n; i++){ // xor of elements in FirstHalfXOR if (i < n / 2) FirstHalfXOR ^= arr[i]; // xor of elements in SecondHalfXOR else SecondHalfXOR ^= arr[i]; } System.out.print(FirstHalfXOR + "," + SecondHalfXOR + "\n"); }// Driver Code public static void main(String[] args) { int arr[] = { 20, 30, 50, 10, 55, 15, 42 }; int N = arr.length; // Function call XOROfElements(arr, N); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find the xor of # the first half elements and # second half elements of an array # Function to find the xor of # the first half elements and # second half elements of an array def XOROfElements(arr, n): FirstHalfXOR = 0; SecondHalfXOR = 0; for i in range(n): # xor of elements in FirstHalfXOR if (i < n // 2): FirstHalfXOR ^= arr[i]; # xor of elements in SecondHalfXOR else: SecondHalfXOR ^= arr[i]; print(FirstHalfXOR,",",SecondHalfXOR);# Driver Code arr = [20, 30, 50, 10, 55, 15, 42]; N = len(arr); # Function call XOROfElements(arr, N); |
C#
// C# program to find the xor of // the first half elements and // second half elements of an array using System;class GFG{// Function to find the xor of // the first half elements and // second half elements of an array static void XOROfElements(int []arr, int n){ int FirstHalfXOR = 0; int SecondHalfXOR = 0; for(int i = 0; i < n; i++) { // xor of elements in FirstHalfXOR if (i < n / 2) FirstHalfXOR ^= arr[i]; // xor of elements in SecondHalfXOR else SecondHalfXOR ^= arr[i]; } Console.Write(FirstHalfXOR + "," + SecondHalfXOR + "\n"); }// Driver Code public static void Main(String[] args) { int []arr = { 20, 30, 50, 10, 55, 15, 42 }; int N = arr.Length; // Function call XOROfElements(arr, N); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to find the xor of // the first half elements and // second half elements of an array // Function to find the xor of // the first half elements and // second half elements of an array function XOROfElements(arr , n) { var FirstHalfXOR = 0; var SecondHalfXOR = 0; for (i = 0; i < n; i++) { // xor of elements in FirstHalfXOR if (i < parseInt(n / 2)) FirstHalfXOR ^= arr[i]; // xor of elements in SecondHalfXOR else SecondHalfXOR ^= arr[i]; } document.write( FirstHalfXOR + ", " + SecondHalfXOR + "\n" ); } // Driver Code var arr = [ 20, 30, 50, 10, 55, 15, 42 ]; var N = arr.length; // Function call XOROfElements(arr, N);// This code contributed by umadevi9616 </script> |
Output:
56, 24
Time complexity: O(N), as we are using a loop to traverse the array.
Auxiliary Space: O(1), as we are not using any extra space.
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