# 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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Improved By : AnkitRai01, Rajput-Ji, vt_m, jit_t

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