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# 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.```

A 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.

1. 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)```
1. Where M is the required X-bit number with all bits set.
2. 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

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## Javascript

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Output:

`11`

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

#### Approach

1. Find the number of bits required to represent the given number.
2. Initialize a variable ‘mask’ to 2^bits-1.
3. XOR the given number with ‘mask’ and store the result in ‘result’.
4. Return ‘result’.

#### Algorithm

1. Initialize a variable ‘num’ with the given number.
2. Initialize a variable ‘result’ to 0.
3. Initialize a variable ‘bits’ to 0.
4. Calculate the number of bits required to represent ‘num’ and store it in ‘bits’.
5. Initialize a variable ‘mask’ to 2^bits-1.
6. XOR ‘num’ with ‘mask’ and store the result in ‘result’.
7. Return ‘result’.

## Python3

 `def` `invert_bits(num):``    ``# Count number of bits``    ``count ``=` `0``    ``temp ``=` `num``    ``while` `temp > ``0``:``        ``count ``+``=` `1``        ``temp >>``=` `1``    ` `    ``# Bitwise NOT of num``    ``# gives a number with all ones``    ``# up to the count of bits``    ``# Bitwise AND with num inverts the bits``    ``return` `(~num) & ((``1` `<< count) ``-` `1``)``num``=``11``print``(invert_bits(num))`

Output

`4`

Time complexity: O(log n) where n is the given number.
Auxiliary Space: O(1)

METHOD 3:Using defaultdict method

APPROACH:

This program uses a defaultdict object to invert the bits of a given input number. The input number is first converted to a binary string, and then each bit of the string is toggled using a defaultdict object. Finally, the inverted bits are converted back to an integer.

ALGORITHM:

1.Convert the input number to a binary string.
2.Create a defaultdict object with int type as the default value.
3.Iterate through the binary string and toggle each bit by setting its value in the defaultdict to the opposite of its original value.
4.Convert the inverted bits back to a binary string and then to an integer using the built-in int() function.
5.Return the inverted integer.

## Python3

 `from` `collections ``import` `defaultdict` `def` `invert_bits(num):``    ``# Convert the number to binary string and remove the '0b' prefix``    ``binary_str ``=` `bin``(num)[``2``:]``    ``# Create a defaultdict with int type as the default value``    ``inverted_bits ``=` `defaultdict(``int``)``    ``# Iterate through the binary string and toggle each bit``    ``for` `i, bit ``in` `enumerate``(binary_str):``        ``inverted_bits[i] ``=` `int``(``not` `int``(bit))``    ``# Convert the inverted bits back to a binary string and then to an integer``    ``inverted_num ``=` `int``(''.join(``str``(inverted_bits[i]) ``for` `i ``in` `range``(``len``(binary_str))), ``2``)``    ``return` `inverted_num` `# Example usage``num ``=` `20``inverted_num ``=` `invert_bits(num)``print``(inverted_num)  ``# Output: 11`

Output

```11
```

Time Complexity:
The time complexity of this program depends on the length of the binary string representation of the input number. The program iterates through the binary string once, which takes O(n) time, where n is the length of the binary string. Converting the inverted bits back to an integer using the int() function takes O(n) time as well. Therefore, the overall time complexity of the program is O(n).

Space Complexity:
The space complexity of this program is also O(n), as the defaultdict object is used to store the inverted bits, and the size of the defaultdict object is proportional to the length of the binary string.

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