Generate an Array with XOR of even length prefix as 0 or 1
Given an integer N, the task is to construct an array of N distinct elements (arr[i] ≤ N+1) such that the bitwise XOR of every prefix having an even length is either 0 or 1.
Examples:
Input: N = 5
Output = 2 3 4 5 6
Explanation: XOR from arr[1] to arr[2] = XOR(2, 3) = 1
XOR from arr[1] to arr[4] = XOR(2, 3, 4, 5) = 0
Input: N = 2
Output: 2 3
Approach: The approach to the problem is based on the following observation
2*k XOR (2*k + 1) = 1 where k ∈ [1, ∞)
The above equation can be proved as shown below:
- 2k is an even number whose LSB is always zero. Adding 1 in this (resulting in 2k+1) will change only one bit of the number (LSB will get transformed from zero to one).
- Now, 2k and 2k+1 differ in only one bit at the 0th position. So, 2*k XOR 2*k+1 = 1.
So, if started from k = 1, and consecutive k‘s are considered the conditions will be satisfied and all the prefixes with even length will have XOR as 1 or 0(when prefix length is divisible by 4. because XOR of even number of 1 will be 0)
Follow the steps mentioned below to implement the above observation:
- Declare a vector to store the answer.
- Run a loop from i = 1 to n/2 and in each iteration:
- store two values in the vector:
- First Value = 2*i.
- Second Value = 2*i + 1.
- If N is odd, insert the last element (N + 1) in the vector because using the above method only an even number of elements can be inserted.
- Return the vector as this is the required array.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
vector< int > construct_arr( int n)
{
vector< int > ans;
for ( int i = 1; i <= n / 2; i++) {
ans.push_back(2 * i);
ans.push_back(2 * i + 1);
}
if ((n % 2) != 0) {
ans.push_back(n + 1);
}
return ans;
}
int main()
{
int N = 5;
vector< int > ans = construct_arr(N);
for ( int i = 0; i < ans.size(); i++)
cout << ans[i] << " " ;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static ArrayList<Integer> construct_arr( int n)
{
ArrayList<Integer> ans = new ArrayList<Integer>();
for ( int i = 1 ; i <= n / 2 ; i++) {
ans.add( 2 *i);
ans.add( 2 *i+ 1 );
}
if ((n % 2 ) != 0 ) {
ans.add(n + 1 );
}
return ans;
}
public static void main(String args[])
{
int N = 5 ;
ArrayList<Integer> ans = construct_arr(N);
for ( int i = 0 ; i < ans.size(); i++)
System.out.print(ans.get(i) + " " );
}
}
|
Python3
def construct_arr(n):
ans = []
for i in range ( 1 , n / / 2 + 1 ):
ans.append( 2 * i)
ans.append( 2 * i + 1 )
if ((n % 2 ) ! = 0 ):
ans.append(n + 1 )
return ans
if __name__ = = "__main__" :
N = 5
ans = construct_arr(N)
for i in range ( 0 , len (ans)):
print (ans[i], end = " " )
|
C#
using System;
using System.Collections.Generic;
class GFG {
static List< int > construct_arr( int n)
{
List< int > ans = new List< int >();
for ( int i = 1; i <= n / 2; i++) {
ans.Add(2 * i);
ans.Add(2 * i + 1);
}
if ((n % 2) != 0) {
ans.Add(n + 1);
}
return ans;
}
public static void Main( string [] args)
{
int N = 5;
List< int > ans = construct_arr(N);
for ( int i = 0; i < ans.Count; i++)
Console.Write(ans[i] + " " );
}
}
|
Javascript
<script>
function construct_arr(n) {
let ans = [];
for (let i = 1; i <= Math.floor(n / 2); i++) {
ans.push(2 * i);
ans.push(2 * i + 1);
}
if ((n % 2) != 0) {
ans.push(n + 1);
}
return ans;
}
let N = 5;
let ans = construct_arr(N);
for (let i = 0; i < ans.length; i++)
document.write(ans[i] + " " )
</script>
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Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
28 Jun, 2022
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