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# Bitwise Operators Examples

Some important tricks with bits that you need to remember

1. Divide by 2:

`x>>=1`

2. Multiply by 2:

`x<<=1`

3. Clear the lowest set bit for x:

`x & (x-1)`

4. Extracting the lowest set bit of x.

`x & ~(x-1)`

5. Clearing all bits from LSB to i’th bit.

`bitmask = ~((1 << i+1 ) - 1);`
`x & = mask ;`

6. Clearing all bits from MSB to the i’th bit.

`bitmask = (1<<i)-1;`
`x & = mask ;`

7. A number x with lowest cleared bit set.

`x | (x + 1)`

8. Extracts the lowest cleared bit of x.

`x | ~(x + 1)`

8. Checking if a number x is a power of 2 or not.

`if (x && !(x & (x-1) ) )==1`
`then x is a power of 2.`

### Example 1:

Checking if the given number is a power of 2.

A very simple solution for this would be to divide the given number by 2 until it is even. If the remaining number is 1 then it is a power of 2 otherwise not a power of 2.

```int PowerOfTwo(int n)
{
if(n==0)
{
return false;
}
else
{
while(n % 2 == 0)
{
n/=2;
}
if(n==1) return true;
else return false;
}
}```
```def PowerOfTwo(n):
if n==0:
return False
else :
while n % 2 == 0 :
n = n / 2
if n==1 :
return True
else :
return False ```
```class PowerOfTwo
{
public static int powerOfTwo(int n)
{
if(n==0)
{
return false;
}
else
{
while(n % 2 == 0)
{
n/=2;
}
if(n==1) return true;
else return false;
}
}
}```

### Example 2:

Finding the x^y in O ( logn ).

This algorithm is one of the most important algorithms in computer science. It is known as the Binary Exponentiation .  The basic idea is that we can represent y in terms of powers of 2 ( Example  y= 13 = 2^3 + 2^2 + 2^1 ) , and now we can write x^y as x^(2^a) * x^(2^b) * ….
Where a, b .. are set bits of y.

An important observation here is that we only need x^(power of 2) , so we can find all x^(2^a) for all a<=63 (as 2^63 is equal to 10^18 and y can only be upto 10^18)  which can be done in O( log y ) and then multiply it to the answer if a is a set bit of y.

As x^y would become very large for some large numbers ( Example 20 ^ 30 ) , instead of finding x^y we would find (x^y) % mod , where mod is a large prime number.

```int power(int x,int y,int mod)
{
int ans=1;
while(y)
{
if(y%2)
{
ans=(ans*x)%mod;
}
y=y>>1;
x=(x*x);
}
return ans;
}```
```def power(x,y,mod):
ans=1;
while y>0 :
if y%2 != 0 :
ans=(ans*x)%mod

y=y>>1
x=x*x
return ans ```
```class BinaryExponentiation
{
public static int power(int x,int y,int mod)
{
int ans=1;
while(y>0)
{
if(y%2 != 0)
{
ans=(ans*x)%mod;
}
y=y>>1;
x=(x*x);
}
return ans;
}
}```

### Examples 3:

Finding the number of set bits of a number.

We are going to use the property ( x & (x-1) ) , which clears the lowest set bit.

```int CountSetBits(int x)
{
int cnt = 0;
while (x)
{
x &= (x-1);
cnt++;
}
return cnt;

}```
```def CountSetbits(x):
cnt=0
while (x) :
x = x & (x-1)
cnt=cnt+1
return cnt ```
```Jclass CountSetBits
{
public static int countSetBits(int x)
{
int cnt=0;
while(x)
{
x= x & (x-1);
cnt++;
}
}
}```

### Example 4:

Checking if the i’th bit is set in the given number x.

When the i’th bit is set in the given number x , then (x&(1<<i)) has to be (1<<i) . ( Here (1<<i) is equal to 2^i ) . This is because (1<<i) has only the i’th bit set. By taking the AND of x with (1<<i) we just check if x has the i’th bit set or not.

```bool ithBitSet(int x , int i)
{
if(x&(1<
```
```def ithBitSet(n):
if x & (1< 0 :
return True
else :
return False```
```class IthBitSet
{
public static int powerOfTwo(int x)
{
if(x&(1<
```

### Example 5:

Finding all subsets of a given array.

We can represent each element in the array as a bit . As we know there are 2^n subsets of an array of size n. So we can iterate i from 0 to 2^n-1 and for each number we can check which bits are set i.e. if the j’th bit of a particular number is 1 then the i’th element would be present in that subset . Let us take one example to understand this more clearly .

Let us say that the array is [ 1 , 2 , 3 ] .

As the size of the array is 3 , we can iterate i from 0 to 2^3-1 i.e from 0 to 7.

```void find_subsets( int arr[] , int n )
{
for ( int i=0 ; i < pow(2,n) ; i++)
{

for(int j=0 ; j < n ; j++)
{
if((i&(1<
```
```def find_subsets( arr , n )

for i in range(0,1<
```
```class Subsets
{
public static void find_subsets( int arr[] , int n )
{
for ( int i=0 ; i < pow(2,n) ; i++)
{

for(int j=0 ; j < n ; j++)
{
if((i&(1<
```

## Serious about Learning Programming ?

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## Bit Manipulation Problems

Bit play
Problem Score Companies Time Status
Number of 1 Bits 200 8:47
Trailing Zeroes 200 14:49
Reverse Bits 225 23:50
Divide Integers 250 68:04
Different Bits Sum Pairwise 300 51:29
Bit tricks
Problem Score Companies Time Status
Min XOR value 200 37:42
Count Total Set Bits 200 63:58
Palindromic Binary Representation 200 60:55
XOR-ing the Subarrays! 200 33:32
Bit array
Problem Score Companies Time Status
Single Number 275 11:53
Single Number II 275 39:22

Problem Score Companies Time Status
Bit Flipping 200 21:31
Swap Bits 150 26:18
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