Given a non-negative number n. The problem is to set the rightmost unset bit in the binary representation of n. If there are no unset bits, then just leave the number as it is.
Examples:
Input : 21 Output : 23 (21)10 = (10101)2 Rightmost unset bit is at position 2(from right) as highlighted in the binary representation of 21. (23)10 = (10111)2 The bit at position 2 has been set. Input : 15 Output : 15
Approach: Following are the steps:
- If n = 0, return 1.
- If all bits of n are set, return n. Refer to this post.
- Else perform bitwise not on the given number(operation equivalent to 1’s complement). Let it be num = ~n.
- Get the position of rightmost set bit of num. Let the position be pos.
- Return (1 << (pos – 1)) | n.
// C++ implementation to set the 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;
} int setRightmostUnsetBit( 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 n;
// position of rightmost unset bit in 'n'
// passing ~n as argument
int pos = getPosOfRightmostSetBit(~n);
// set the bit at position 'pos'
return ((1 << (pos - 1)) | n);
} // Driver program to test above int main()
{ int n = 21;
cout << setRightmostUnsetBit(n);
return 0;
} |
// Java implementation to set // the 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 ;
} static int setRightmostUnsetBit( 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 n;
// position of rightmost unset bit in 'n'
// passing ~n as argument
int pos = getPosOfRightmostSetBit(~n);
// set the bit at position 'pos'
return (( 1 << (pos - 1 )) | n);
} // Driver code public static void main(String arg[]) {
int n = 21 ;
System.out.print(setRightmostUnsetBit(n));
} } // This code is contributed by Anant Agarwal. |
# Python3 implementation to # set the rightmost unset bit import math
# function to find the position # of rightmost set bit def getPosOfRightmostSetBit(n):
return int (math.log2(n& - n) + 1 )
def setRightmostUnsetBit(n):
# if n = 0, return 1
if (n = = 0 ):
return 1
# if all bits of 'n' are set
if ((n & (n + 1 )) = = 0 ):
return n
# position of rightmost unset bit in 'n'
# passing ~n as argument
pos = getPosOfRightmostSetBit(~n)
# set the bit at position 'pos'
return (( 1 << (pos - 1 )) | n)
# Driver code n = 21
print (setRightmostUnsetBit(n))
# This code is contributed # by Anant Agarwal. |
// C# implementation to set // the 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;
} static int setRightmostUnsetBit( 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 n;
// Position of rightmost unset bit in 'n'
// passing ~n as argument
int pos = getPosOfRightmostSetBit(~n);
// Set the bit at position 'pos'
return ((1 << (pos - 1)) | n);
} // Driver code public static void Main(String []arg)
{ int n = 21;
Console.Write(setRightmostUnsetBit(n));
} } // This code is contributed by shivanisinghss2110 |
<script> // JavaScript implementation to set // the rightmost unset bit // function to find the position
// of rightmost set bit
function getPosOfRightmostSetBit(n)
{
return ((Math.log10(n & -n)) / (Math.log10(2))) + 1;
}
function setRightmostUnsetBit(n)
{
// if n = 0, return 1
if (n == 0)
return 1;
// if all bits of 'n' are set
if ((n & (n + 1)) == 0)
return n;
// position of rightmost unset bit in 'n'
// passing ~n as argument
let pos = getPosOfRightmostSetBit(~n);
// set the bit at position 'pos'
return ((1 << (pos - 1)) | n);
}
// Driver code
let n = 21;
document.write(setRightmostUnsetBit(n));
// This code is contributed by unknown2108 </script> |
Output:
23
Alternate Implementation
The idea is to use Integer.toBinaryString()
// C++ program to implement the approach #include <bits/stdc++.h> using namespace std;
// function to binary string and remove // the leading zeroes string getBinary( int n)
{ string binary = bitset<32>(n).to_string();
binary.erase(0, binary.find_first_not_of( '0' ));
return binary;
} void setMostRightUnset( int a)
{ // will get a number with all set bits except the
// first set bit
int x = a ^ (a - 1);
cout << getBinary(x) << endl;
// We reduce it to the number with single 1's on
// the position of first set bit in given number
x = x & a;
cout << getBinary(x) << endl;
// Move x on right by one shift to make OR
// operation and make first rightest unset bit 1
x = x >> 1;
int b = a | x;
cout << "before setting bit " << getBinary(a) << endl;
cout << "after setting bit " << getBinary(b) << endl;
} int main()
{ int a = 21;
setMostRightUnset(a);
return 0;
} // This code is contributed by phasing17 |
import java.io.*;
import java.util.Scanner;
public class SetMostRightUnsetBit {
public static void main(String args[])
{
int a = 21 ;
setMostRightUnset(a);
}
private static void setMostRightUnset( int a)
{
// will get a number with all set bits except the
// first set bit
int x = a ^ (a - 1 );
System.out.println(Integer.toBinaryString(x));
// We reduce it to the number with single 1's on
// the position of first set bit in given number
x = x & a;
System.out.println(Integer.toBinaryString(x));
// Move x on right by one shift to make OR
// operation and make first rightest unset bit 1
x = x >> 1 ;
int b = a | x;
System.out.println( "before setting bit " +
Integer.toBinaryString(a));
System.out.println( "after setting bit " +
Integer.toBinaryString(b));
}
} |
def setMostRightUnset(a):
# Will get a number with all set
# bits except the first set bit
x = a ^ (a - 1 )
print ( bin (x)[ 2 :])
# We reduce it to the number with
# single 1's on the position of
# first set bit in given number
x = x & a
print ( bin (x)[ 2 :])
# Move x on right by one shift to
# make OR operation and make first
# rightest unset bit 1
x = x >> 1
b = a | x
print ( "before setting bit " , bin (a)[ 2 :])
print ( "after setting bit " , bin (b)[ 2 :])
# Driver Code if __name__ = = '__main__' :
a = 21
setMostRightUnset(a)
# This code is contributed by mohit kumar 29 |
using System;
class SetMostRightUnsetBit{
// Driver Code public static void Main(String []args)
{ int a = 21;
setMostRightUnset(a);
} private static void setMostRightUnset( int a)
{ // will get a number with all set bits
// except the first set bit
int x = a ^ (a - 1);
Console.WriteLine(Convert.ToString(x));
// We reduce it to the number with single 1's on
// the position of first set bit in given number
x = x & a;
Console.WriteLine(Convert.ToString(x));
// Move x on right by one shift to make OR
// operation and make first rightest unset bit 1
x = x >> 1;
int b = a | x;
Console.WriteLine( "before setting bit " +
Convert.ToString(a, 2));
Console.WriteLine( "after setting bit " +
Convert.ToString(b, 2));
} } // This code is contributed by shivanisinghss2110 |
<script> function setMostRightUnset(a)
{
// will get a number with all set bits except the
// first set bit
let x = a ^ (a - 1);
document.write((x >>> 0).toString(2)+ "<br>" );
// We reduce it to the number with single 1's on
// the position of first set bit in given number
x = x & a;
document.write((x >>> 0).toString(2)+ "<br>" );
// Move x on right by one shift to make OR
// operation and make first rightest unset bit 1
x = x >> 1;
let b = a | x;
document.write( "before setting bit " +
(a >>> 0).toString(2)+ "<br>" );
document.write( "after setting bit " +
(b >>> 0).toString(2)+ "<br>" );
}
let a = 21;
setMostRightUnset(a);
// This code is contributed by patel2127 </script> |
Output:
1 1 before setting bit 10101 after setting bit 10101
Another Approach:
The idea is x|(x+1) sets the lowest(rightmost) unset bit but it may set a leading zero bit if the given number doesn’t contain any unset bits , for example
if all bits set 15 is 1111 = 00001111 16 is 10000 = 00010000 _________ x|(x+1) 00011111 on performing x|(x+1) it converts to 00011111 but here we should only consider only 1111 and return 1111 as there are no unset bits. else 13 is 1101 = 00001101 14 is 1110 = 00001110 _________ x|(x+1) 00001111 on performing x|(x+1) it converts to 00011111 which is required number.
So we can eliminate this type of redundant operation by checking if there are any zeroes in the given number. We observe only number whose all bits set are of form 2k -1 have no unset bits in them so by catching these number we can perform our idea.
// C++ implementation to set the rightmost unset bit #include <bits/stdc++.h> using namespace std;
int setRightmostUnsetBit( int n)
{ // if all bits of 'n' are set
// the number is of form 2^k -1 return n
if (!(n & (n + 1)))
return n;
// else
return n | (n + 1);
} // Driver program to test above int main()
{ int n = 21;
cout << setRightmostUnsetBit(n);
return 0;
} // This code is contributed by Kasina Dheeraj. |
// Java implementation to set // the rightmost unset bit class GFG {
static int setRightmostUnsetBit( int n)
{
// if all bits of 'n' are set
// the number is of form 2^k -1 return n
if ((n & (n + 1 )) == 0 )
return n;
// else
return n | (n + 1 );
}
// Driver code
public static void main(String arg[])
{
int n = 21 ;
System.out.print(setRightmostUnsetBit(n));
}
} // This code is contributed by Kasina Dheeraj. |
# Python3 implementation to # set the rightmost unset bit import math
def setRightmostUnsetBit(n):
# if all bits of 'n' are set
# the number is of form 2^k -1 return n
if ((n & (n + 1 )) = = 0 ):
return n
# else
return n | (n + 1 )
# Driver code n = 21
print (setRightmostUnsetBit(n))
# This code is contributed by Kasina Dheeraj. |
// C# implementation to set // the rightmost unset bit using System;
class GFG {
static int setRightmostUnsetBit( int n)
{
// if all bits of 'n' are set
// the number is of form 2^k -1 return n
if ((n & (n + 1)) == 0)
return n;
// else
return n | (n + 1);
}
// Driver code
public static void Main(String[] arg)
{
int n = 21;
Console.Write(setRightmostUnsetBit(n));
}
} // This code is contributed by Kasina Dheeraj. |
// JavaScript implementation to set the rightmost unset bit function setRightmostUnsetBit(n)
{ // if all bits of 'n' are set
// the number is of form 2^k -1 return n
if (!(n & (n + 1)))
return n;
// else
return n | (n + 1);
} // Driver program to test above let n = 21; console.log(setRightmostUnsetBit(n)); // This code is contributed by phasing17 |
Output:
23
Time complexity: O(1)
Space Complexity: O(1)