Skip to content
Related Articles

Related Articles

Improve Article

Get the position of rightmost unset bit

  • Difficulty Level : Easy
  • Last Updated : 05 Apr, 2021

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

This article is contributed by Ayush Jauhari. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :