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Get the position of rightmost unset bit

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Given a non-negative number n. Find the position of rightmost unset bit in the binary representation of n, considering the last bit at position 1, 2nd last bit at position 2 and so on. If no 0’s are there in the binary representation of n. then print “-1”.
Examples: 
 

Input : n = 9
Output : 2
(9)10 = (1001)2
The position of rightmost unset bit in the binary
representation of 9 is 2.

Input : n = 32
Output : 1


 


Approach: Following are the steps:
 

  1. If n = 0, return 1.
  2. If all bits of n are set, return -1. Refer this post.
  3. Else perform bitwise not on the given number(operation equivalent to 1’s complement). Let it be num = ~n.
  4. Get the position of rightmost set bit of num. This will be the position of rightmost unset bit of n.


 

C++

// C++ implementation to get the position of rightmost unset bit
#include <bits/stdc++.h>
 
using namespace std;
 
// function to find the position
// of rightmost set bit
int getPosOfRightmostSetBit(int n)
{
    return log2(n&-n)+1;
}
 
// function to get the position of rightmost unset bit
int getPosOfRightMostUnsetBit(int n)
{
    // if n = 0, return 1
    if (n == 0)
        return 1;
     
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)   
        return -1;
     
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    return getPosOfRightmostSetBit(~n);       
}
 
// Driver program to test above
int main()
{
    int n = 9;
    cout << getPosOfRightMostUnsetBit(n);
    return 0;
}

                    

Java

// Java implementation to get the
// position of rightmost unset bit
 
class GFG {
     
// function to find the position
// of rightmost set bit
static int getPosOfRightmostSetBit(int n)
{
    return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}
 
// function to get the position
// of rightmost unset bit
static int getPosOfRightMostUnsetBit(int n) {
     
    // if n = 0, return 1
    if (n == 0)
    return 1;
 
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)
    return -1;
 
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    return getPosOfRightmostSetBit(~n);
}
 
// Driver code
public static void main(String arg[])
{
    int n = 9;
    System.out.print(getPosOfRightMostUnsetBit(n));
}
}
 
// This code is contributed by Anant Agarwal.

                    

Python3

# Python3 implementation to get the position
# of rightmost unset bit
 
# import library
import math as m
  
# function to find the position
# of rightmost set bit
def getPosOfRightmostSetBit(n):
     
    return (m.log(((n & - n) + 1),2))
 
  
# function to get the position ot rightmost unset bit
def getPosOfRightMostUnsetBit(n):
     
    # if n = 0, return 1
    if (n == 0):
        return 1
      
    # if all bits of 'n' are set
    if ((n & (n + 1)) == 0):
        return -1
      
    # position of rightmost unset bit in 'n'
    # passing ~n as argument
    return getPosOfRightmostSetBit(~n)   
 
  
# Driver program to test above
n = 13;
ans = getPosOfRightMostUnsetBit(n)
 
#rounding the final answer
print (round(ans))
 
# This code is contributed by Saloni Gupta.

                    

C#

// C# implementation to get the
// position of rightmost unset bit
using System;
 
class GFG
{
    // function to find the position
    // of rightmost set bit
    static int getPosOfRightmostSetBit(int n)
    {
        return (int)((Math.Log10(n & -n)) / Math.Log10(2)) + 1;
    }
     
    // function to get the position
    // of rightmost unset bit
    static int getPosOfRightMostUnsetBit(int n) {
         
        // if n = 0, return 1
        if (n == 0)
        return 1;
     
        // if all bits of 'n' are set
        if ((n & (n + 1)) == 0)
        return -1;
     
        // position of rightmost unset bit in 'n'
        // passing ~n as argument
        return getPosOfRightmostSetBit(~n);
    }
     
    // Driver code
    public static void Main()
    {
        int n = 9;
        Console.Write(getPosOfRightMostUnsetBit(n));
    }
}
 
// This code is contributed by Sam007

                    

PHP

<?php
// PHP implementation to get the
// position of rightmost unset bit
 
// function to find the position
// of rightmost set bit
function getPosOfRightmostSetBit( $n)
{
    return ceil(log($n &- $n) + 1);
}
 
// function to get the position
// of rightmost unset bit
function getPosOfRightMostUnsetBit( $n)
{
    // if n = 0, return 1
    if ($n == 0)
        return 1;
     
    // if all bits of 'n' are set
    if (($n & ($n + 1)) == 0)
        return -1;
     
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    return getPosOfRightmostSetBit(~$n);    
}
 
    // Driver Code
    $n = 9;
    echo getPosOfRightMostUnsetBit($n);
     
// This code is contributed by anuj_67.
?>

                    

Javascript

<script>
// JavaScript implementation to get the position of rightmost unset bit
 
// function to find the position
// of rightmost set bit
function getPosOfRightmostSetBit(n)
{
    return Math.log2(n&-n)+1;
}
 
// function to get the position of rightmost unset bit
function getPosOfRightMostUnsetBit(n)
{
 
    // if n = 0, return 1
    if (n == 0)
        return 1;
     
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)   
        return -1;
     
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    return getPosOfRightmostSetBit(~n);       
}
 
// Driver program to test above
    let n = 9;
    document.write(getPosOfRightMostUnsetBit(n));
 
// This code is contributed by Manoj.
</script>

                    

Output: 
 

2

Time Complexity – O(1)

Space Complexity – O(1)




 



Last Updated : 22 Jun, 2022
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