I have to flip all bits in a binary representation of an integer. Given:
10101
The output should be
01010
What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like int flipBits(int n);, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.
int findComplement(int num) {
int i = 0, ans = 0;
while(num) {
if(not (num & 1)) {
ans += (1 << i);
}
i += 1;
num >>= 1;
}
return ans;
}
Binary 10101 == Decimal 21
Flipped Binary 01010 == Decimal 10
One liner (in Javascript - You could convert to your favorite programming language )
10 == ~21 & (1 << (Math.floor(Math.log2(21))+1)) - 1
Explanation:
10 == ~21 & mask
mask : For filtering out all the leading bits before the significant bits count (nBits - see below)
How to calculate the significant bit counts ?
Math.floor(Math.log2(21))+1 => Returns how many significant bits are there (nBits)
Ex:
0000000001 returns 1
0001000001 returns 7
0000010101 returns 5
(1 << nBits) - 1 => 1111111111.....nBits times = mask
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 . For example: Number: Given Number Value : A number with all bits set in a 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 a 32-bit integer.
int setBitNumber(int n)
{
n |= n>>1;
n |= n>>2;
n |= n>>4;
n |= n>>8;
n |= n>>16;
n = n + 1;
return (n >> 1);
}