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.
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