Replace every array element by Bitwise Xor of previous and next element
Given an array of integers, replace every element with xor of previous and next elements with following exceptions.
- First element is replaced by xor of first and second.
- Last element is replaced by xor of last and second last.
Examples:
Input: arr[] = { 2, 3, 4, 5, 6}
Output: 1 6 6 2 3
We get the following array as {2^3, 2^4, 3^5, 4^6, 5^6}
Input: arr[] = { 1, 2, 1, 5}
Output: 3, 0, 7, 4
We get the following array as {1^2, 1^1, 2^5, 1^5}
A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.
C++
#include <iostream>
using namespace std;
void ReplaceElements( int arr[], int n)
{
int newarr[n];
for ( int i=0;i<n;i++){
newarr[i]=arr[i];}
arr[0]=arr[0]^arr[1];
for ( int i=1;i<n-1;i++){
arr[i]=newarr[i-1]^newarr[i+1];}
arr[n-1]=newarr[n-1]^newarr[n-2];
}
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
public class Main {
public static void replaceElements( int [] arr, int n) {
int [] newarr = new int [n];
for ( int i = 0 ; i < n; i++) {
newarr[i] = arr[i];
}
arr[ 0 ] = arr[ 0 ] ^ arr[ 1 ];
for ( int i = 1 ; i < n - 1 ; i++) {
arr[i] = newarr[i - 1 ] ^ newarr[i + 1 ];
}
arr[n - 1 ] = newarr[n - 1 ] ^ newarr[n - 2 ];
}
public static void main(String[] args) {
int [] arr = { 2 , 3 , 4 , 5 , 6 };
int n = arr.length;
replaceElements(arr, n);
for ( int i = 0 ; i < n; i++) {
System.out.print(arr[i] + " " );
}
}
}
|
Python3
def replaceElements(arr, n):
newarr = [ 0 ] * n
for i in range (n):
newarr[i] = arr[i]
arr[ 0 ] = arr[ 0 ] ^ arr[ 1 ]
for i in range ( 1 , n - 1 ):
arr[i] = newarr[i - 1 ] ^ newarr[i + 1 ]
arr[n - 1 ] = newarr[n - 1 ] ^ newarr[n - 2 ]
arr = [ 2 , 3 , 4 , 5 , 6 ]
n = len (arr)
replaceElements(arr, n)
for i in range (n):
print (arr[i], end = " " )
|
Javascript
function ReplaceElements(arr, n) {
let newarr = [...arr];
arr[0] = arr[0] ^ arr[1];
for (let i = 1; i < n - 1; i++) {
arr[i] = newarr[i - 1] ^ newarr[i + 1];
}
arr[n - 1] = newarr[n - 1] ^ newarr[n - 2];
}
let arr = [2, 3, 4, 5, 6];
let n = arr.length;
ReplaceElements(arr, n);
for (let i = 0; i < n; i++) {
console.log(arr[i] + " " );
}
|
C#
using System;
class MainClass {
static void ReplaceElements( int [] arr, int n)
{
int [] newarr = new int [n];
for ( int i = 0; i < n; i++) {
newarr[i] = arr[i];
}
arr[0]
= arr[0] ^ arr[1];
for ( int i = 1; i < n - 1;
i++) {
arr[i]
= newarr[i - 1]
^ newarr[i + 1];
}
arr[n - 1]
= newarr[n - 1]
^ newarr[n - 2];
}
public static void Main()
{
int [] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
}
|
Time Complexity: O(n)
Auxiliary Space: O(n)
An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Xor the previous element using the extra variable and the next element to get each element.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
void ReplaceElements( int arr[], int n)
{
if (n <= 1)
return ;
int prev = arr[0];
arr[0] = arr[0] ^ arr[1];
for ( int i = 1; i < n - 1; i++) {
int curr = arr[i];
arr[i] = prev ^ arr[i + 1];
prev = curr;
}
arr[n - 1] = prev ^ arr[n - 1];
}
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java .io.*;
class GFG
{
static void ReplaceElements( int [] arr,
int n)
{
if (n <= 1 )
return ;
int prev = arr[ 0 ];
arr[ 0 ] = arr[ 0 ] ^ arr[ 1 ];
for ( int i = 1 ; i < n - 1 ; i++)
{
int curr = arr[i];
arr[i] = prev ^ arr[i + 1 ];
prev = curr;
}
arr[n - 1 ] = prev ^ arr[n - 1 ];
}
public static void main(String[] args)
{
int [] arr = { 2 , 3 , 4 , 5 , 6 };
int n = arr.length;
ReplaceElements(arr, n);
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
}
|
Python3
def ReplaceElements(arr, n):
if n < = 1 :
return
prev = arr[ 0 ]
arr[ 0 ] = arr[ 0 ] ^ arr[ 1 ]
for i in range ( 1 , n - 1 ):
curr = arr[i]
arr[i] = prev ^ arr[i + 1 ]
prev = curr
arr[n - 1 ] = prev ^ arr[n - 1 ]
arr = [ 2 , 3 , 4 , 5 , 6 ]
n = len (arr)
ReplaceElements(arr, n)
for i in range (n):
print (arr[i], end = " " )
|
C#
using System;
class GFG
{
static void ReplaceElements( int [] arr,
int n)
{
if (n <= 1)
return ;
int prev = arr[0];
arr[0] = arr[0] ^ arr[1];
for ( int i = 1; i < n - 1; i++)
{
int curr = arr[i];
arr[i] = prev ^ arr[i + 1];
prev = curr;
}
arr[n - 1] = prev ^ arr[n - 1];
}
public static void Main()
{
int [] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
}
|
PHP
<?php
function ReplaceElements(& $arr , $n )
{
if ( $n <= 1)
return ;
$prev = $arr [0];
$arr [0] = $arr [0] ^ $arr [1];
for ( $i = 1; $i < $n - 1; $i ++)
{
$curr = $arr [ $i ];
$arr [ $i ] = $prev ^ $arr [ $i + 1];
$prev = $curr ;
}
$arr [ $n - 1] = $prev ^ $arr [ $n - 1];
}
$arr = array ( 2, 3, 4, 5, 6 );
$n = sizeof( $arr );
ReplaceElements( $arr , $n );
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ] . " " ;
?>
|
Javascript
<script>
function ReplaceElements(arr, n)
{
if (n <= 1)
return ;
let prev = arr[0];
arr[0] = arr[0] ^ arr[1];
for (let i = 1; i < n - 1; i++)
{
let curr = arr[i];
arr[i] = prev ^ arr[i + 1];
prev = curr;
}
arr[n - 1] = prev ^ arr[n - 1];
}
let arr = [ 2, 3, 4, 5, 6 ];
let n = arr.length;
ReplaceElements(arr, n);
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
</script>
|
Time complexity: O(N), where N is the number of elements in the given array.
Auxiliary space: O(1)
Last Updated :
06 Apr, 2023
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