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) :
(n % 4 == 0)
(n % 4 == 1)
(n % 4 == 2)
n + 1;
- 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?
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++.
number = 0b011;
cout << number;
- 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.
Number : Given Number
Value : A number with all bits set in given number.
Flipped number = Value – Number.
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.
n |= n>>1;
n |= n>>2;
n |= n>>4;
n |= n>>8;
n |= n>>16;
n = n + 1;
(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.
This article is contributed by Sanchit Garg 1. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.