# Bits manipulation (Important tactics)

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2022

## CPP

 `// Direct XOR of all numbers from 1 to n``int` `computeXOR(``int` `n)``{``    ``if` `(n % 4 == 0)``        ``return` `n;``    ``if` `(n % 4 == 1)``        ``return` `1;``    ``if` `(n % 4 == 2)``        ``return` `n + 1;``    ``else``        ``return` `0;``}`

## Javascript

 ``

```Input: 6
Output: 7```
1. Refer Compute XOR from 1 to n for details.
2. We can quickly calculate the total number of combinations with numbers smaller than or equal to a number whose sum and XOR are equal. Instead of using looping (Brute force method), we can directly find it by a mathematical trick i.e.

```// Refer Equal Sum and XOR for details.
Answer = pow(2, count of zero bits)```

## CPP

 `//  Function to check if x is power of 2``bool` `isPowerOfTwo(``int` `x)``{``     ``// First x in the below expression is``     ``// for  the case when x is 0``     ``return` `x && (!(x & (x - 1)));``}`

Refer check if a number is power of two for details.

#### 3. Find XOR of all subsets of a set

We can do it in O(1) time. The answer is always 0 if the given set has more than one element. For sets with a single element, the answer is the value of single element.

Refer XOR of the XOR’s of all subsets for details.

#### 4. Find number of leading, trailing zeroes and number of 1’s

We can quickly find number of leading, trailing zeroes and number of 1’s in a binary code of an integer in C++ using GCC. It can be done by using inbuilt function i.e.

```  Number of leading zeroes: __builtin_clz(x)
Number of trailing zeroes : __builtin_ctz(x)
Number of 1-bits: __builtin_popcount(x) ```

Refer GCC inbuilt functions for details.

## CPP

 `// Conversion into Binary code//``#include ``using` `namespace` `std;` `int` `main()``{``    ``auto` `number = 0b011;``    ``cout << number;``    ``return` `0;``}`

`Output: 3`

#### 6. The Quickest way to swap two numbers:

```a ^= b;
b ^= a;
a ^= b;```

Refer swap two numbers for details.

#### 7. Simple approach to flip the bits of a number:

It can be done in a simple way, just simply subtract the number from the value obtained when all the bits are equal to 1.
For example:

```Number : Given Number
Value  : A number with all bits set in given number.
Flipped number = Value – Number.

Example :
Number = 23,
Binary form: 10111;
After flipping digits number will be: 01000;
Value: 11111 = 31;```

#### 8. Finding the most significant set bit

We can find the most significant set bit in O(1) time for a fixed size integer. For example below code is for 32-bit integer.

## C++

 `int` `setBitNumber(``int` `n)``{     ``    ``// Below steps set bits after``    ``// MSB (including MSB)` `    ``// Suppose n is 273 (binary``    ``// is 100010001). It does following``    ``// 100010001 | 010001000 = 110011001``    ``n |= n>>1;` `    ``// This makes sure 4 bits``    ``// (From MSB and including MSB)``    ``// are set. It does following``    ``// 110011001 | 001100110 = 111111111``    ``n |= n>>2;  ` `    ``n |= n>>4; ``    ``n |= n>>8;``    ``n |= n>>16;``    ` `    ``// Increment n by 1 so that``    ``// there is only one set bit``    ``// which is just before original``    ``// MSB. n now becomes 1000000000``    ``n = n + 1;` `    ``// Return original MSB after shifting.``    ``// n now becomes 100000000``    ``return` `(n >> 1);``}`

Refer Find most significant set bit of a number for details.

#### 9.Check if a number has bits in alternate pattern

We can quickly check if bits in a number are in alternate pattern (like 101010). We compute n ^ (n >> 1). If n has an alternate pattern, then n ^ (n >> 1) operation will produce a number having set bits only. ‘^’ is a bitwise XOR operation.

## C++

 `// function to check if a number has bits in alternate``// pattern``bool` `bitsAreInAltOrder(unsigned ``int` `n)``{``    ``unsigned ``int` `num = n ^ (n >> 1);``    ``// to check if all bits are set in 'num'``    ``return` `allBitsAreSet(num);``}`

Refer check if a number has bits in alternate pattern for details.

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