Find array whose elements are XOR of adjacent elements in given array
Last Updated :
14 Sep, 2021
Given an array arr[] consisting of N integers, the task is to re-construct an array arr[] such that the values in arr[] are obtained by doing XOR of the adjacent elements in the array. Print the array elements.
Examples:
Input: arr[ ] = {10, 11, 1, 2, 3}
Output: 1 10 3 1 3
Explanation:
At index 0, arr[0] xor arr[1] = 1
At index 1, arr[1] xor arr[2] = 10
At index 2, arr[2] xor arr[3] = 3
…
At index 4, No element is left So, it will remain as it is.
New Array will be {1, 10, 3, 1, 3}
Input: arr[ ] = {5, 9, 7, 6}
Output: 12 14 1 6
Explanation:
At index 0, arr[0] xor arr[1] = 12
At index 1, arr[1] xor arr[2] = 14
At index 2, arr[2] xor arr[3] = 1
At index 3, No element is left So, it will remain as it is.
New Array will be {12, 14, 1, 6}
Approach: The main idea to solve the given problem is to perform the following steps:
- Traverse the given array arr[] from the 0th index to (N – 2)th index.
- For each element arr[i] at ith position calculate arr[i] ^ arr[i+1] and store it at position i.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int * game_with_number( int arr[], int n)
{
for ( int i = 0; i < n - 1; i++) {
arr[i] = arr[i] ^ arr[i + 1];
}
return arr;
}
void print( int arr[], int n)
{
for ( int i = 0; i < n; i++) {
cout << arr[i] << " " ;
}
}
int main()
{
int arr[] = { 10, 11, 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
int * new_arr = game_with_number(arr, n);
print(new_arr, n);
}
|
Java
import java.io.*;
class GFG{
static int [] game_with_number( int arr[], int n)
{
for ( int i = 0 ; i < n - 1 ; i++)
{
arr[i] = arr[i] ^ arr[i + 1 ];
}
return arr;
}
static void print( int arr[], int n)
{
for ( int i = 0 ; i < n; i++)
{
System.out.print(arr[i] + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 10 , 11 , 1 , 2 , 3 };
int n = arr.length;
int [] new_arr = game_with_number(arr, n);
print(new_arr, n);
}
}
|
Python3
def game_with_number(arr, n):
for i in range (n - 1 ):
arr[i] = arr[i] ^ arr[i + 1 ]
return arr
def printt(arr, n):
print ( * arr)
if __name__ = = '__main__' :
arr = [ 10 , 11 , 1 , 2 , 3 ]
n = len (arr)
new_arr = game_with_number(arr, n);
printt(new_arr, n)
|
C#
using System;
class GFG{
static int [] game_with_number( int [] arr, int n)
{
for ( int i = 0; i < n - 1; i++)
{
arr[i] = arr[i] ^ arr[i + 1];
}
return arr;
}
static void print( int [] arr, int n)
{
for ( int i = 0; i < n; i++)
{
Console.Write(arr[i] + " " );
}
}
public static void Main()
{
int [] arr = { 10, 11, 1, 2, 3 };
int n = arr.Length;
int [] new_arr = game_with_number(arr, n);
print(new_arr, n);
}
}
|
Javascript
<script>
function game_with_number(arr,n)
{
for (let i = 0; i < n - 1; i++)
{
arr[i] = arr[i] ^ arr[i + 1];
}
return arr;
}
function print(arr,n)
{
for (let i = 0; i < n; i++) {
document.write(arr[i]+ " " );
}
}
let arr = [10, 11, 1, 2, 3 ];
let n = arr.length;
let new_arr = game_with_number(arr, n);
print(new_arr, n);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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