- Compute XOR from 1 to n (direct method) :
// Direct XOR of all numbers from 1 to n
(n % 4 == 0)
(n % 4 == 1)
(n % 4 == 2)
n + 1;
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 with 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?
// Function to check if x is power of 2
// First x in the below expression is
// for the case when x is 0
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 given set has more than one elements. For set with single element, the answer is 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++.
// Conversion into Binary code//
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 by 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. 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 cocde is for 32 bit integer.
// 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
(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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