# Turn off the rightmost set bit

Write a program that unsets the rightmost set bit of an integer.
Examples :

```Input:  12 (00...01100)
Output: 8 (00...01000)

Input:  7 (00...00111)
Output: 6 (00...00110)```

Let the input number be n. n-1 would have all the bits flipped after the rightmost set bit (including the set bit). So, doing n&(n-1) would give us the required result.

So, now let us see how n – 1 is flipping all the bits to the right (including the rightmost set bit also) of the n
Taking n = 12, so (n – 1) = 11,

n can be written like (n = (n – 1) + 1), so now we can think of this problem as Adding 1 to Any Number (refer this article for better understanding)
Binary representation of (n-1) = 11 = 1011, so now lets make n from (n-1), which can be done by adding a 1 to (n-1)
On adding 1 to any number X, all bits to the right of  rightmost 0 (including the rightmost zero) gets flipped
(n-1) = 1011

(n-1) + 1 = 1100 (all the bits to the right of rightmost zero (including rightmost zero) got flipped)
Since we have flipped the rightmost zero, so now, rightmost zero is now flipped to rightmost 1 (aka the rightmost set bit of n) and all bits before rightmost 0 are the same as before
X = 010 . . . . . 0 (rightmost zero) 111

X + 1 = 010 . . . . . 1 (rightmost one) 0 0 0

Example :

X = 71 = Think of it as n – 1

Binary Representation of X = 1000111

X + 1 = 72 = Think of it as n

Binary Representation of (X+1) = 1001000
Observation :

1. All the bits to the left of rightmost 0 (excluding rightmost 0) in X (thinking it as n – 1) are same as in to the left of the rightmost 1(excluding rightmost 1)  in X + 1 (thinking of it as n)

2. All the bits to the right of rightmost 0 (including rightmost 0) in X (thinking it as n – 1) are different as in to the right of the rightmost 1 (including rightmost 1) in X + 1 (thinking of it as n)

So bitwise AND of left part of X (till rightmost 0, excluding rightmost 0) and left part of X + 1 (till rightmost 1, excluding rightmost 1) will give the required answer,  bitwise AND right part of X (from rightmost 0) and right part of X + 1 (from rightmost 1 (rightmost set bit)) will result in

## C++

 `#include ` `using` `namespace` `std;`   `// unsets the rightmost set bit ` `// of n and returns the result ` `int` `fun(unsigned ``int` `n) ` `{ ` `    ``return` `n & (n - 1); ` `} `   `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 7; ` `    ``cout<<``"The number after unsetting the"``; ` `    ``cout<<``" rightmost set bit "``<

## C

 `#include `   `// unsets the rightmost set bit` `// of n and returns the result` `int` `fun(unsigned ``int` `n)` `{` `    ``return` `n & (n - 1);` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 7;` `    ``printf``(``"The number after unsetting the"``);` `    ``printf``(``" rightmost set bit %d"``, fun(n));` `    ``return` `0;` `}`

## Java

 `// Java program to unset the` `// rightmost set bit of an integer.` `import` `java.io.*;` `class` `GFG {`   `    ``/* unsets the rightmost set bit` `     ``of n and returns the result */` `    ``static` `int` `fun(``int` `n)` `    ``{` `        ``return` `n & (n - ``1``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String arg[])` `    ``{` `        ``int` `n = ``7``;` `        ``System.out.print(``"The number after unsetting "` `                         ``+ ``"the rightmost set bit "` `+ fun(n));` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# unsets the rightmost set bit` `# of n and returns the result ` `def` `fun(n):`   `     ``return` `n & (n``-``1``)`   `# Driver code`   `n ``=` `7` `print``(``"The number after unsetting the rightmost set bit"``, fun(n))`   `# This code is contributed` `# by Anant Agarwal.`

## C#

 `// C# program to unset the` `// rightmost set bit of an integer.` `using` `System;`   `class` `GFG {`   `    ``/* unsets the rightmost set bit` `    ``of n and returns the result */` `    ``static` `int` `fun(``int` `n)` `    ``{` `        ``return` `n & (n - 1);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 7;` `        ``Console.Write(``"The number after unsetting "` `                      ``+ ``"the rightmost set bit "` `+ fun(n));` `    ``}` `}`   `// This code is contributed by Sam007`

## Javascript

 ``

## PHP

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Output

`The number after unsetting the rightmost set bit 6`

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Approach:

The rightmost set bit can be switched off by subtracting the LSB from the number.

The LSB of a number can be obtained using (n & (-n)), therefore the number with the rightmost set bit of n switched off is equal to n – (n & (-n));

## C++

 `//C++ program to implement the approach`   `#include ` `using` `namespace` `std;`   `// unsets the rightmost set bit ` `// of n and returns the result ` `int` `fun(unsigned ``int` `n) ` `{ ` `    ``return`  `n - (n & (-n)); ` `} `   `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 7; ` `    ``cout<<``"The number after unsetting the"``; ` `    ``cout<<``" rightmost set bit: "``<

## Java

 `// Java program to implement the approach` `import` `java.io.*;` `class` `GFG` `{`   `  ``// unsets the rightmost set bit` `  ``// of n and returns the result` `  ``static` `int` `fun(``int` `n)` `  ``{` `    ``return` `n - (n & (-n));` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int` `n = ``7``;` `    ``System.out.print(``"The number after unsetting the"``);` `    ``System.out.print(``" rightmost set bit: "` `+ fun(n));` `  ``}` `}`   `// This code is contributed by phasing17`

## Python3

 `# Python3 program to implement the approach`   `# unsets the rightmost set bit` `# of n and returns the result` `def` `fun(n):` `    ``return` `n ``-` `(n & (``-``n))`   `# Driver Code` `n ``=` `7` `print``(``"The number after unsetting the rightmost set bit:"``, fun(n))`   `# This code is contributed by phasing17`

## C#

 `// C# program to implement the approach` `using` `System;`   `class` `GFG {`   `  ``// unsets the rightmost set bit` `  ``// of n and returns the result` `  ``static` `int` `fun(``int` `n) { ``return` `n - (n & (-n)); }`   `  ``// Driver Code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``int` `n = 7;` `    ``Console.Write(``"The number after unsetting the"``);` `    ``Console.Write(``" rightmost set bit: "` `+ fun(n));` `  ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 `// JavaScript program to implement the approach`   `// unsets the rightmost set bit ` `// of n and returns the result ` `function` `fun(n) ` `{ ` `    ``return`  `n - (n & (-n)); ` `} `   `// Driver Code ` `let n = 7; ` `console.log(``"The number after unsetting the rightmost set bit:"``, fun(n)); `   `// This code is contributed by phasing17`

Output

`The number after unsetting the rightmost set bit: 6`

Time Complexity: O(1)
Auxiliary Space: O(1)

Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem

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