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
Approach: This is a constructive problem and may contain multiple solutions. Follow the below steps to generate the required sequence:
- Take first N – 3 elements as part of the sequence i.e. 1, 2, 3, 4, …, (N – 3)
- 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 sequencevoid 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 codeint main(){ int n = 4, x = 3; Findseq(n, x); return 0;} |
Java
// Java implementation of the approachimport 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 approachusing 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 sequencefunction 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 sequencefunction 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), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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