Replace every array element by Bitwise Xor of previous and next element
Last Updated :
06 Apr, 2023
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++
// C++ program to update every array element with
// sum of previous and next numbers in array
#include <iostream>
using namespace std;
void ReplaceElements(int arr[], int n)
{
int newarr[n];//created a new array to store the elements.
for(int i=0;i<n;i++){
newarr[i]=arr[i];}
arr[0]=arr[0]^arr[1];// changed the first element
for(int i=1;i<n-1;i++){// iterated from second element to last second
arr[i]=newarr[i-1]^newarr[i+1];}// element and changed every element according to question.
arr[n-1]=newarr[n-1]^newarr[n-2];// changed the last element of the array.
}
// Driver program
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
//This code is contributed by Naveen Gujjar from Haryana
public class Main {
public static void replaceElements(int[] arr, int n) {
int[] newarr = new int[n]; // created a new array to store the elements.
for (int i = 0; i < n; i++) {
newarr[i] = arr[i];
}
arr[0] = arr[0] ^ arr[1]; // changed the first element
for (int i = 1; i < n - 1; i++) { // iterated from second element to last second
arr[i] = newarr[i - 1] ^ newarr[i + 1]; // element and changed every element according to the question.
}
arr[n - 1] = newarr[n - 1] ^ newarr[n - 2]; // changed the last element of the array.
}
// Driver program
public static void main(String[] args) {
int[] arr = { 2, 3, 4, 5, 6 };
int n = arr.length;
replaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
}
# Function to update every array element with sum of previous and next numbers in array
def replaceElements(arr, n):
newarr = [0] * n # created a new array to store the elements
for i in range(n):
newarr[i] = arr[i]
arr[0] = arr[0] ^ arr[1] # changed the first element
for i in range(1, n-1): # iterated from second element to last second element and changed every element according to question
arr[i] = newarr[i-1] ^ newarr[i+1]
# changed the last element of the array
arr[n-1] = newarr[n-1] ^ newarr[n-2]
# Driver program
arr = [2, 3, 4, 5, 6]
n = len(arr)
replaceElements(arr, n)
# Print the modified array
for i in range(n):
print(arr[i], end=" ")
// JS program to update every array element with
// sum of previous and next numbers in array
function ReplaceElements(arr, n) {
let newarr = [...arr]; // created a new array to store the elements.
arr[0] = arr[0] ^ arr[1]; // changed the first element
for (let i = 1; i < n - 1; i++) {
// iterated from second element to last second element
arr[i] = newarr[i - 1] ^ newarr[i + 1]; // element and changed every element according to question.
}
arr[n - 1] = newarr[n - 1] ^ newarr[n - 2]; // changed the last element of the array.
}
// Driver program
let arr = [2, 3, 4, 5, 6];
let n = arr.length;
ReplaceElements(arr, n);
// Print the modified array
for (let i = 0; i < n; i++) {
console.log(arr[i] + " ");
}
// C# program to update every array element with
// sum of previous and next numbers in array
using System;
class MainClass {
static void ReplaceElements(int[] arr, int n)
{
int[] newarr = new int[n]; // created a new array to
// store the elements.
for (int i = 0; i < n; i++) {
newarr[i] = arr[i];
}
arr[0]
= arr[0] ^ arr[1]; // changed the first element
for (int i = 1; i < n - 1;
i++) { // iterated from second element to last
// second
arr[i]
= newarr[i - 1]
^ newarr[i + 1]; // element and changed
// every element according
// to question.
}
arr[n - 1]
= newarr[n - 1]
^ newarr[n - 2]; // changed the last element
// of the array.
}
public static void Main()
{
int[] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
ReplaceElements(arr, n);
// Print the modified array
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++
// C++ program to update every array element with
// sum of previous and next numbers in array
#include <iostream>
using namespace std;
void ReplaceElements(int arr[], int n)
{
// Nothing to do when array size is 1
if (n <= 1)
return;
// store current value of arr[0] and update it
int prev = arr[0];
arr[0] = arr[0] ^ arr[1];
// Update rest of the array elements
for (int i = 1; i < n - 1; i++) {
// Store current value of next interaction
int curr = arr[i];
// Update current value using previous value
arr[i] = prev ^ arr[i + 1];
// Update previous value
prev = curr;
}
// Update last array element separately
arr[n - 1] = prev ^ arr[n - 1];
}
// Driver program
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
// Java program to update every array
// element with sum of previous and
// next numbers in array
import java .io.*;
class GFG
{
static void ReplaceElements(int[] arr,
int n)
{
// Nothing to do when array size is 1
if (n <= 1)
return;
// store current value of arr[0]
// and update it
int prev = arr[0];
arr[0] = arr[0] ^ arr[1];
// Update rest of the array elements
for (int i = 1; i < n - 1; i++)
{
// Store current value of
// next interaction
int curr = arr[i];
// Update current value using
// previous value
arr[i] = prev ^ arr[i + 1];
// Update previous value
prev = curr;
}
// Update last array element separately
arr[n - 1] = prev ^ arr[n - 1];
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 3, 4, 5, 6 };
int n = arr.length;
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
}
// This code is contributed
// by anuj_67..
# Python3 program to update every
# array element with sum of previous
# and next numbers in array
def ReplaceElements(arr, n):
# Nothing to do when array
# size is 1
if n <= 1:
return
# store current value of arr[0]
# and update it
prev = arr[0]
arr[0] = arr[0] ^ arr[1]
# Update rest of the array elements
for i in range(1, n - 1):
# Store current value of
# next interaction
curr = arr[i]
# Update current value using
# previous value
arr[i] = prev ^ arr[i + 1]
# Update previous value
prev = curr
# Update last array element separately
arr[n - 1] = prev ^ arr[n - 1]
# Driver Code
arr = [2, 3, 4, 5, 6]
n = len(arr)
ReplaceElements(arr, n)
for i in range(n):
print(arr[i], end = " ")
# This code is contributed
# by Shrikant13
// C# program to update every array
// element with sum of previous and
// next numbers in array
using System;
class GFG
{
static void ReplaceElements(int[] arr,
int n)
{
// Nothing to do when array size is 1
if (n <= 1)
return;
// store current value of arr[0]
// and update it
int prev = arr[0];
arr[0] = arr[0] ^ arr[1];
// Update rest of the array elements
for (int i = 1; i < n - 1; i++)
{
// Store current value of
// next interaction
int curr = arr[i];
// Update current value using
// previous value
arr[i] = prev ^ arr[i + 1];
// Update previous value
prev = curr;
}
// Update last array element separately
arr[n - 1] = prev ^ arr[n - 1];
}
// Driver Code
public static void Main()
{
int[] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
}
// This code is contributed
// by Akanskha Rai(Abby_akku)
<?php
// PHP program to update every array
// element with sum of previous and
// next numbers in array
function ReplaceElements(&$arr, $n)
{
// Nothing to do when array size is 1
if ($n <= 1)
return;
// store current value of arr[0]
// and update it
$prev = $arr[0];
$arr[0] = $arr[0] ^ $arr[1];
// Update rest of the array elements
for ($i = 1; $i < $n - 1; $i++)
{
// Store current value of next
// interaction
$curr = $arr[$i];
// Update current value using
// previous value
$arr[$i] = $prev ^ $arr[$i + 1];
// Update previous value
$prev = $curr;
}
// Update last array element separately
$arr[$n - 1] = $prev ^ $arr[$n - 1];
}
// Driver Code
$arr = array( 2, 3, 4, 5, 6 );
$n = sizeof($arr);
ReplaceElements($arr, $n);
// Print the modified array
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
// This code is contributed by ita_c
?>
<script>
// Javascript program to update every array
// element with sum of previous and next
// numbers in array
function ReplaceElements(arr, n)
{
// Nothing to do when array size is 1
if (n <= 1)
return;
// Store current value of arr[0] and update it
let prev = arr[0];
arr[0] = arr[0] ^ arr[1];
// Update rest of the array elements
for(let i = 1; i < n - 1; i++)
{
// Store current value of next interaction
let curr = arr[i];
// Update current value using previous value
arr[i] = prev ^ arr[i + 1];
// Update previous value
prev = curr;
}
// Update last array element separately
arr[n - 1] = prev ^ arr[n - 1];
}
// Driver code
let arr = [ 2, 3, 4, 5, 6 ];
let n = arr.length;
ReplaceElements(arr, n);
// Print the modified array
for(let i = 0; i < n; i++)
document.write(arr[i] + " ");
// This code is contributed by subhammahato348
</script>
Time complexity: O(N), where N is the number of elements in the given array.
Auxiliary space: O(1)
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