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Find smallest number n such that n XOR n+1 equals to given k.
  • Difficulty Level : Hard
  • Last Updated : 10 May, 2018

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 n
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 program
int main()
{
    int k = 31;
    cout << xorCalc(k);
    return 0;
}

Java




// Java to find n such that XOR of n and n+1
// is equals to given n
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 (String[] args)
    {
        int k = 31;
          
        System.out.println(xorCalc(k));
    }
}
  
// This code is contributed by Anant Agarwal.

Python3




# python to find n such that
# XOR of n and n+1 is equals
# to given n
  
# function to return the 
# required n
def 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 program
k = 31
print(int(xorCalc(k)))
  
# This code is contributed
# by Sam007

C#




// C# to find n such that XOR
// of n and n+1 is equals to
// given n
using 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 n
function 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.
?>


Output:
15

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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