You are given a positive number k, we need to find a positive integer n, such that XOR of n and n+1 is equal to k. If no such n exist then print -1.
Examples:
Input : 3 Output : 1 Input : 7 Output : 3 Input : 6 Output : -1
Below are two cases when we do n XOR (n+1) for a number n.
Case 1 : n is even. Last bit of n is 0 and last bit of (n+1) is 1. Rest of the bits are same in both. So XOR would always be 1 if n is even.
Case : n is odd Last bit in n is 1. And in n+1, last bit is 0. But in this case there may be more bits which differ due to carry. The carry continues to propagate to left till we find first 0 bit. So n XOR n+1 will we 2^i-1 where i is the position of first 0 bit in n from left. So, we can say that if k is of form 2^i-1 then we will have our answer as k/2.
Finally our steps are:
If we have k=1, answer = 2 [We need smallest positive n] Else If k is of form 2^i-1, answer = k/2, else, answer = -1
C++
// CPP to find n such that XOR of n and n+1 is equals to// given n#include <bits/stdc++.h>using namespace std;// function to return the required nint xorCalc(int k){ if (k == 1) return 2; // if k is of form 2^i-1 if (((k + 1) & k) == 0) return k / 2; return -1;}// driver programint main(){ int k = 31; cout << xorCalc(k); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C to find n such that XOR of n and n+1 is equals to given n#include <stdio.h>// function to return the required nint xorCalc(int k){ if (k == 1) return 2; // if k is of form 2^i-1 if (((k + 1) & k) == 0) return k / 2; return -1;}// driver programint main(){ int k = 31; printf("%d",xorCalc(k)); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java to find n such that XOR of n and n+1// is equals to given nclass GFG { // function to return the required n static int xorCalc(int k) { if (k == 1) return 2; // if k is of form 2^i-1 if (((k + 1) & k) == 0) return k / 2; return 1; } // Driver code public static void main(String[] args) { int k = 31; System.out.println(xorCalc(k)); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# python to find n such that# XOR of n and n+1 is equals# to given n# function to return the # required ndef xorCalc(k): if (k == 1): return 2 # if k is of form 2^i-1 if (((k + 1) & k) == 0): return k // 2 return -1;# driver programk = 31print(int(xorCalc(k)))# This code is contributed# by Shushant Kumar |
C#
// C# to find n such that XOR// of n and n+1 is equals to// given nusing System;class GFG { // function to return the required // n static int xorCalc(int k) { if (k == 1) return 2; // if k is of form 2^i-1 if (((k + 1) & k) == 0) return k / 2; return 1; } // Driver code public static void Main () { int k = 31; Console.WriteLine(xorCalc(k)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP to find n such // that XOR of n and n+1// is equals to given n// function to return // the required nfunction xorCalc($k){ if ($k == 1) return 2; // if k is of form 2^i-1 if ((($k + 1) & $k) == 0) return floor($k / 2); return 1;}// Driver Code$k = 31;echo xorCalc($k);// This code is contributed by vt_m.?> |
Javascript
<script>// Javascript to find n such that XOR of n and n+1// is equals to given n// function to return the required nfunction xorCalc(k){ if (k == 1) return 2; // if k is of form 2^i-1 if (((k + 1) & k) == 0) return parseInt(k / 2); return 1;}// driver programvar k = 31;document.write( xorCalc(k));// This code is contributed by itsok.</script> |
Output:
15
Time Complexity : O(1)
Auxiliary Space : O(1)
This article is contributed by Shivam Pradhan (anuj_charm). If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
