# Program to invert bits of a number Efficiently

Given a non-negative integer N. The task is to invert the bits of the number N and print the decimal equivalent of the number obtained after inverting the bits.

Note: Leading 0’s are not being considered.

Examples:

```Input : 11
Output : 4
(11)10 = (1011)2
After inverting the bits, we get:
(0100)2 = (4)10.

Input : 20
Output : 11
(20)10 = (10100)2.
After inverting the bits, we get:
(01011)2 = (11)10.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The similar problem is already discussed in Invert actual bits of a number.

In this article, an efficient approach using bitwise operators is discussed. Below is the step by step algorithm to solve the problem:

1. Calculate the total number of bits in the given number. This can be done by calculating:
```X = log2N
```

Where, N is the given number and X is the total number of bits of N.

2. The next step is to generate a number with X bits and all bits set. That is, 11111….X-times. This can be done by calculating:
```Step-1: M = 1 << X
Step-2: M = M | (M-1)
```

Where M is the required X-bit number with all bits set.

3. The final step is to calculate the bit-wise XOR of M with N, which will be our answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to invert actual bits ` `// of a number. ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to invert bits of a number ` `int` `invertBits(``int` `n) ` `{ ` `    ``// Calculate number of bits of N-1; ` `    ``int` `x = log2(n) ; ` ` `  `    ``int` `m = 1 << x; ` ` `  `    ``m = m | m - 1; ` ` `  `    ``n = n ^ m; ` ` `  `    ``return` `n; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 20; ` ` `  `    ``cout << invertBits(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to invert  ` `// actual bits of a number.  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `// Function to invert  ` `// bits of a number  ` `static` `int` `invertBits(``int` `n)  ` `{  ` `    ``// Calculate number of bits of N-1;  ` `    ``int` `x = (``int``)(Math.log(n) /  ` `                  ``Math.log(``2``)) ;  ` ` `  `    ``int` `m = ``1` `<< x;  ` ` `  `    ``m = m | m - ``1``;  ` ` `  `    ``n = n ^ m;  ` ` `  `    ``return` `n;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `n = ``20``;  ` ` `  `    ``System.out.print(invertBits(n));  ` `} ` `}  ` ` `  `// This code is contributed by Smitha `

## Python3

 `# Python3 program to invert actual ` `# bits of a number. ` `import` `math ` ` `  `# Function to invert bits of a number ` `def` `invertBits(n): ` `     `  `    ``# Calculate number of bits of N-1 ` `    ``x ``=` `int``(math.log(n, ``2``)) ` ` `  `    ``m ``=` `1` `<< x ` ` `  `    ``m ``=` `m | m ``-` `1` ` `  `    ``n ``=` `n ^ m ` ` `  `    ``return` `n ` ` `  `# Driver code ` `n ``=` `20` ` `  `print``(invertBits(n)) ` ` `  `# This code is contributed 29AjayKumar `

## C#

 `// C# program to invert  ` `// actual bits of a number.  ` `using` `System; ` ` `  ` `  `  `  `public` `class` `GFG ` `{ ` `// Function to invert  ` `// bits of a number  ` `static` `int` `invertBits(``int` `n)  ` `{  ` `    ``// Calculate number of bits of N-1;  ` `    ``int` `x = (``int``)(Math.Log(n) /  ` `                  ``Math.Log(2)) ;  ` `  `  `    ``int` `m = 1 << x;  ` `  `  `    ``m = m | m - 1;  ` `  `  `    ``n = n ^ m;  ` `  `  `    ``return` `n;  ` `}  ` `  `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `n = 20;  ` `  `  `    ``Console.Write(invertBits(n));  ` `} ` `}  ` `  `  `// This code is contributed by Subhadeep `

## PHP

 ` `

Output:

```11
```

Time Complexity: O(log2n)
Auxiliary Space: O(1)

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