 Practice
Resources
Contests
Online IDE
New
Free Mock
Events New Scaler
Practice
Improve your coding skills with our resources
Contests
Compete in popular contests with top coders Events
Attend free live masterclass hosted by top tech professionals
New
Scaler
Explore Offerings by SCALER Go to Problems

# 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 ?

Learn this and a lot more with Scaler Academy's industry vetted curriculum which covers Data Structures & Algorithms in depth.

## Bit Manipulation Problems

Bit play
Problem Score Companies Time Status
Number of 1 Bits 200 8:47
Trailing Zeroes 200 14:54
Reverse Bits 225 23:50
Divide Integers 250 68:08
Different Bits Sum Pairwise 300 51:30
Bit tricks
Problem Score Companies Time Status
Min XOR value 200 37:42
Count Total Set Bits 200 65:06
Palindromic Binary Representation 200 63:15
XOR-ing the Subarrays! 200 34:14
Bit array
Problem Score Companies Time Status
Single Number 275 11:53
Single Number II 275 39:22 Excel at your interview with Masterclasses Know More   Certificate included What will you Learn? Free Mock Assessment
Fill up the details for personalised experience.
Phone Number *
OTP will be sent to this number for verification
+1 *
+1
Change Number
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
2028
*Enter the expected year of graduation if you're student
Current Employer
Company Name
College/University Name
Job Title
Job Title
Software Development Engineer (Backend)
Software Development Engineer (Frontend)
Software Development Engineer (Full Stack)
Data Scientist
Android Engineer
iOS Engineer
Devops Engineer
Support Engineer
Research Engineer
Engineering Intern
QA Engineer
Co-founder
SDET
Product Manager
Product Designer
Backend Architect
Program Manager
Release Engineer 