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Create a sequence whose XOR of elements is y

  • Last Updated : 27 Apr, 2021

Given two integers N and Y, the task is to generate a sequence of N distinct non-negative integers whose bitwise-XOR of all the elements of this generated sequence is equal to Y i.e. A1 ^ A2 ^ A3 ^ ….. ^ AN = Y where ^ denotes bitwise XOR. if no such sequence is possible then print -1.
Examples: 
 

Input: N = 4, Y = 3 
Output: 1 131072 131074 0 
(1 ^ 131072 ^ 131074 ^ 0) = 3 and all four elements are distinct.
Input: N = 10, Y = 6 
Output: 1 2 3 4 5 6 7 131072 131078 0 
 

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Approach: This is a constructive problem and may contain multiple solutions. Follow the below steps to generate the required sequence: 
 

  1. Take first N – 3 elements as part of the sequence i.e. 1, 2, 3, 4, …, (N – 3)
  2. Let the XOR of the chosen elements be x and num be an integer that has not been chosen yet. Now there are two cases: 
    • If x = y then we can add num, num * 2 and (num ^ (num * 2)) to the last 3 remaining numbers because num ^ (num * 2) ^ (num ^ (num * 2)) = 0 and x ^ 0 = x
    • If x != y then we can add 0, num and (num ^ x ^ y) because 0 ^ num ^ (num ^ x ^ y) = x ^ y and x ^ x ^ y = y

Note: Sequence is not possible when N = 2 and Y = 0 because this condition can only be satisfied by two equal numbers which is not allowed.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find and print
// the required sequence
void Findseq(int n, int x)
{
    const int pw1 = (1 << 17);
    const int pw2 = (1 << 18);
 
    // Base case
    if (n == 1)
        cout << x << endl;
 
    // Not allowed case
    else if (n == 2 && x == 0)
        cout << "-1" << endl;
    else if (n == 2)
        cout << x << " "
             << "0" << endl;
    else {
        int i;
        int ans = 0;
 
        // XOR of first N - 3 elements
        for (i = 1; i <= n - 3; i++) {
            cout << i << " ";
            ans = ans ^ i;
        }
 
        // Case 1: Add three integers whose XOR is 0
        if (ans == x)
            cout << pw1 + pw2 << " "
                 << pw1 << " " << pw2 << endl;
 
        // Case 2: Add three integers
        // whose XOR is equal to ans
        else
            cout << pw1 << " " << ((pw1 ^ x) ^ ans)
                 << " 0 " << endl;
    }
}
 
// Driver code
int main()
{
    int n = 4, x = 3;
    Findseq(n, x);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to find and print
    // the required sequence
    static void Findseq(int n, int x)
    {
        int pw1 = 1 << 17;
        int pw2 = (1 << 18);
 
        // Base case
        if (n == 1) {
            System.out.println(x);
        }
 
        // Not allowed case
        else if (n == 2 && x == 0) {
            System.out.println("-1");
        }
        else if (n == 2) {
            System.out.println(x + " "
                               + "");
        }
        else {
            int i;
            int ans = 0;
 
            // XOR of first N - 3 elements
            for (i = 1; i <= n - 3; i++) {
                System.out.print(i + " ");
                ans = ans ^ i;
            }
 
            // Case 1: Add three integers whose XOR is 0
            if (ans == x) {
                System.out.println(pw1 + pw2 + " " + pw1 + " " + pw2);
            }
 
            // Case 2: Add three integers
            // whose XOR is equal to ans
            else {
                System.out.println(pw1 + " " + ((pw1 ^ x) ^ ans)
                                   + " 0 ");
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 4, x = 3;
        Findseq(n, x);
    }
}
 
// This code contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to find and print
# the required sequence
def Findseq(n, x) :
     
    pw1 = (1 << 17);
    pw2 = (1 << 18);
 
    # Base case
    if (n == 1) :
        print(x);
 
    # Not allowed case
    elif (n == 2 and x == 0) :
        print("-1");
         
    elif (n == 2) :
        print(x, " ", "0");
         
    else :
     
        ans = 0;
 
        # XOR of first N - 3 elements
        for i in range(1, n - 2) :
            print(i, end = " ");
            ans = ans ^ i;
         
        # Case 1: Add three integers whose XOR is 0
        if (ans == x) :
            print(pw1 + pw2, " ", pw1, " ", pw2);
 
        # Case 2: Add three integers
        # whose XOR is equal to ans
        else :
            print(pw1, " ", ((pw1 ^ x) ^ ans), " 0 ");
 
# Driver code
if __name__ == "__main__" :
     
    n = 4; x = 3;
    Findseq(n, x);
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to find and print
    // the required sequence
    static void Findseq(int n, int x)
    {
        int pw1 = 1 << 17;
        int pw2 = (1 << 18);
 
        // Base case
        if (n == 1)
        {
            Console.WriteLine(x);
        }
 
        // Not allowed case
        else if (n == 2 && x == 0)
        {
            Console.WriteLine("-1");
        }
        else if (n == 2)
        {
            Console.WriteLine(x + " "
                            + "");
        }
        else
        {
            int i;
            int ans = 0;
 
            // XOR of first N - 3 elements
            for (i = 1; i <= n - 3; i++)
            {
                Console.Write(i + " ");
                ans = ans ^ i;
            }
 
            // Case 1: Add three integers whose XOR is 0
            if (ans == x)
            {
                Console.WriteLine(pw1 + pw2 + " " + pw1 + " " + pw2);
            }
 
            // Case 2: Add three integers
            // whose XOR is equal to ans
            else
            {
                Console.WriteLine(pw1 + " " + ((pw1 ^ x) ^ ans)
                                + " 0 ");
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4, x = 3;
        Findseq(n, x);
    }
}
 
// This code contributed by anuj_67..

PHP




<?php
// PHP implementation of the approach
// Function to find and print
// the required sequence
function Findseq($n, $x)
{
    $pw1 = (1 << 17);
    $pw2 = (1 << 18);
 
    // Base case
    if ($n == 1)
        echo $x ,"\n";
 
    // Not allowed case
    else if ($n == 2 && $x == 0)
        echo "-1" ,"\n";
    else if ($n == 2)
        echo $x ," ",
            "0","\n";
    else
    {
        $i;
        $ans = 0;
 
        // XOR of first N - 3 elements
        for ($i = 1; $i <= $n - 3; $i++)
        {
            echo $i , " ";
            $ans = $ans ^ $i;
        }
 
        // Case 1: Add three integers whose XOR is 0
        if ($ans == $x)
            echo ($pw1 + $pw2) , " ",
                $pw1 ," " , $pw2,"\n";
 
        // Case 2: Add three integers
        // whose XOR is equal to ans
        else
            echo $pw1 , " " ,(($pw1 ^ $x) ^ $ans),
                " 0 " ,"\n";
    }
}
 
// Driver code
 
    $n = 4;
    $x = 3;
    Findseq($n, $x);
 
// This code is contributed BY @Tushil..
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to find and print
// the required sequence
function Findseq(n, x)
{
    let pw1 = (1 << 17);
    let pw2 = (1 << 18);
 
    // Base case
    if (n == 1)
        document.write(x + "<br>");
 
    // Not allowed case
    else if (n == 2 && x == 0)
        document.write("-1<br>");
    else if (n == 2)
        document.write(x + " "
             + "0" + "<br>");
    else {
        let i;
        let ans = 0;
 
        // XOR of first N - 3 elements
        for (i = 1; i <= n - 3; i++) {
            document.write(i + " ");
            ans = ans ^ i;
        }
 
        // Case 1: Add three integers whose XOR is 0
        if (ans == x)
            document.write((pw1 + pw2) + " "
                 + pw1 + " " <+ pw2 + "<br>");
 
        // Case 2: Add three integers
        // whose XOR is equal to ans
        else
            document.write(pw1 + " " + ((pw1 ^ x) ^ ans)
                 + " 0 " + "<br>");
    }
}
 
// Driver code
    let n = 4, x = 3;
    Findseq(n, x)
 
</script>
Output: 
1 131072 131074 0

 

Time Complexity: O(N)
 




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