# Bits manipulation (Important tactics)

Prerequisites : Bitwise operators in C, Bitwise Hacks for Competitive Programming, Bit Tricks for Competitive Programming

- Compute XOR from 1 to n (direct method) :

## 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

`<script>` `// Direct XOR of all numbers from 1 to n` `function` `computeXOR(n)` `{` ` ` `if` `(n % 4 == 0)` ` ` `return` `n;` ` ` `if` `(n % 4 == 1)` ` ` `return` `1;` ` ` `if` `(n % 4 == 2)` ` ` `return` `n + 1;` ` ` `else` ` ` `return` `0;` `}` `// This code is contributed by Shubham Singh` `</script>` |

Input: 6 Output: 7

- Refer Compute XOR from 1 to n for details.
- 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)

- How to know if a number is a power of 2?

## 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.
- 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.
- 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.
- Convert binary code directly into an integer in C++.

## CPP

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

Output: 3

- The Quickest way to swap two numbers:

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

- Refer swap two numbers for details.

- 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;

- 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.

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

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