Generate an Array with XOR of even length prefix as 0 or 1

Last Updated : 28 Jun, 2022

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++

// C++ code to implement the approach  #include <bits/stdc++.h> using namespace std;  // Function to construct the array 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 is odd insert the last element     if ((n % 2) != 0) {         ans.push_back(n + 1);     }     return ans; }  // Driver code int main() {     int N = 5;      // Function call     vector<int> ans = construct_arr(N);      // Print the resultant array     for (int i = 0; i < ans.size(); i++)         cout << ans[i] << " ";      return 0; }

Java

/*package whatever //do not write package name here */ import java.io.*; import java.util.*;  class GFG  {    // Java program for the above approach    // Function to construct the array   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 is odd insert the last element     if ((n % 2) != 0) {       ans.add(n + 1);     }     return ans;   }    // Driver Code   public static void main(String args[])   {     int N = 5;      // Function call     ArrayList<Integer> ans = construct_arr(N);      // Print the resultant array     for (int i = 0; i < ans.size(); i++)       System.out.print(ans.get(i) + " ");    } }  // This code is contributed by shinjanpatra.

Python3

# python code to implement the approach  # Function to construct the array def construct_arr(n):      ans = []     for i in range(1, n//2+1):         ans.append(2 * i)         ans.append(2 * i + 1)      # If n is odd insert the last element     if ((n % 2) != 0):         ans.append(n + 1)      return ans   # Driver code if __name__ == "__main__":      N = 5      # Function call     ans = construct_arr(N)      # Print the resultant array     for i in range(0, len(ans)):         print(ans[i], end=" ")      # This code is contributed by rakeshsahni

// C# program for the above appraochh using System; using System.Collections.Generic;  class GFG {    // Function to construct the array   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 is odd insert the last element     if ((n % 2) != 0) {       ans.Add(n + 1);     }     return ans;   }    // Driver Code   public static void Main(string[] args)   {     int N = 5;      // Function call     List<int> ans = construct_arr(N);      // Print the resultant array     for (int i = 0; i < ans.Count; i++)       Console.Write(ans[i] + " ");   } }  // This code is contributed by phasing17

JavaScript

<script>         // JavaScript code for the above approach          // Function to construct the array         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 is odd insert the last element             if ((n % 2) != 0) {                 ans.push(n + 1);             }             return ans;         }          // Driver code          let N = 5;          // Function call         let ans = construct_arr(N);          // Print the resultant array         for (let i = 0; i < ans.length; i++)             document.write(ans[i] + " ")      // This code is contributed by Potta Lokesh     </script>