Given an array arr and a number K, find the new array formed by performing XOR of the corresponding element from the given array with the given number K.
Examples:
Input: arr[] = { 2, 4, 1, 3, 5 }, K = 5
Output: 7 1 4 6 0
Explanation:
2 XOR 5 = 7
4 XOR 5 = 1
1 XOR 5 = 4
3 XOR 5 = 6
5 XOR 5 = 0
Input: arr[] = { 4, 75, 45, 42 }, K = 2
Output: 6 73 47 40
Approach:
- Traverse the given array.
- Then calculate the XOR of each element with K.
- Then store it as the element at that index in the output array.
- Print the updated array.
Below is the implementation of the above approach.
CPP
// C++ program to find XOR of every element// of an array with a given number K#include <bits/stdc++.h>using namespace std;// Function to construct new arrayvoid constructXORArray(int A[], int n, int K){ int B[n]; // Traverse the array and // compute XOR with K for (int i = 0; i < n; i++) B[i] = A[i] ^ K; // Print new array for (int i = 0; i < n; i++) cout << B[i] << " "; cout << endl;}// Driver codeint main(){ int A[] = { 2, 4, 1, 3, 5 }; int K = 5; int n = sizeof(A) / sizeof(A[0]); constructXORArray(A, n, K); int B[] = { 4, 75, 45, 42 }; K = 2; n = sizeof(B) / sizeof(B[0]); constructXORArray(B, n, K); return 0;} |
Java
// Java program to find XOR of every element// of an array with a given number Kimport java.util.*;class GFG{// Function to construct new arraystatic void constructXORArray(int A[], int n, int K){ int[] B = new int[n]; // Traverse the array and // compute XOR with K for (int i = 0; i < n; i++) B[i] = A[i] ^ K; // Print new array for (int i = 0; i < n; i++) System.out.print( B[i] +" "); System.out.println();}// Driver code public static void main(String args[]) { int A[] = { 2, 4, 1, 3, 5 }; int K = 5; int n = A.length; constructXORArray(A, n, K); int B[] = { 4, 75, 45, 42 }; K = 2; n = B.length; constructXORArray(B, n, K);}}// This code is contributed by shivanisinghss2110 |
Python3
# Python program to find XOR of every element# of an array with a given number K# Function to construct new arraydef constructXORArray(A, n, K): B = [0]*n; # Traverse the array and # compute XOR with K for i in range(n): B[i] = A[i] ^ K; # Print new array for i in range(n): print(B[i], end=" "); print();# Driver codeif __name__ == '__main__': A = [ 2, 4, 1, 3, 5 ]; K = 5; n = len(A); constructXORArray(A, n, K); B = [ 4, 75, 45, 42 ]; K = 2; n = len(B); constructXORArray(B, n, K);# This code contributed by sapnasingh4991 |
C#
// C# program to find XOR of every element// of an array with a given number Kusing System;class GFG{ // Function to construct new arraystatic void constructXORArray(int []A, int n, int K){ int[] B = new int[n]; // Traverse the array and // compute XOR with K for (int i = 0; i < n; i++) B[i] = A[i] ^ K; // Print new array for (int i = 0; i < n; i++) Console.Write( B[i] +" "); Console.WriteLine();} // Driver code public static void Main(String []args) { int []A = { 2, 4, 1, 3, 5 }; int K = 5; int n = A.Length; constructXORArray(A, n, K); int []B = { 4, 75, 45, 42 }; K = 2; n = B.Length; constructXORArray(B, n, K);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to find XOR of every element // of an array with a given number K // Function to construct new array function constructXORArray(A, n, K) { let B = new Array(n); // Traverse the array and // compute XOR with K for (let i = 0; i < n; i++) B[i] = A[i] ^ K; // Print new array for (let i = 0; i < n; i++) document.write(B[i] + " "); document.write("</br>") } let A = [ 2, 4, 1, 3, 5 ]; let K = 5; let n = A.length; constructXORArray(A, n, K); let B = [ 4, 75, 45, 42 ]; K = 2; n = B.length; constructXORArray(B, n, K); // This code is contributed by divyeshrabadiya07.</script> |
Output:
7 1 4 6 0 6 73 47 40
Time Complexity: O(n)
Auxiliary Space: O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
