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# Set the rightmost unset bit

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:

1. If n = 0, return 1.
2. If all bits of n are set, return n. Refer to 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. Let the position be pos.
5. Return (1 << (pos – 1)) | n.

## C++

 `// C++ implementation to set the rightmost unset bit``#include ``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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`23`

Alternate Implementation
The idea is to use Integer.toBinaryString()

## C++

 `// C++ program to implement the approach``#include ``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`

## Java

 `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));``    ``}``}`

## Python3

 `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`

## C#

 `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`

## Javascript

 ``

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++

 `// C++ implementation to set the rightmost unset bit``#include ``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

 `// 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

 `# 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#

 `// 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

 `// 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)

This article is contributed by Ayush Jauhari and Kasina Dheeraj. If you like GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.
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