Create a sequence whose XOR of elements is y

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 which 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 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



Output:

1 131072 131074 0

Time Complexity: O(N)

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