Bitwise Hacks for Competitive Programming

Prerequisite: It is recommended to refer Interesting facts about Bitwise Operators 

How to set a bit in the number 'num':

If we want to set a bit at nth position in the number 'num', it can be done using the 'OR' operator( | ).  

• First, we left shift '1' to n position via (1<<n)
• Then, use the 'OR' operator to set the bit at that position. 'OR' operator is used because it will set the bit even if the bit is unset previously in the binary representation of the number 'num'.

Note: If the bit would be already set then it would remain unchanged.

6

Time Complexity: O(1)
Auxiliary Space: O(1)

We have passed the parameter by 'call by reference' to make permanent changes in the number.

2. How to unset/clear a bit at n'th position in the number 'num' : 

Suppose we want to unset a bit at nth position in number 'num' then we have to do this with the help of 'AND' (&) operator.

• First, we left shift '1' to n position via (1<<n) then we use bitwise NOT operator '~' to unset this shifted '1'.
• Now after clearing this left shifted '1' i.e making it to '0' we will 'AND'(&) with the number 'num' that will unset bit at nth position.

5

Time Complexity: O(1)
Auxiliary Space: O(1)

3.  Toggling a bit at nth position :

Toggling means to turn bit 'on'(1) if it was 'off'(0) and to turn 'off'(0) if it was 'on'(1) previously. We will be using the 'XOR' operator here which is this '^'. The reason behind the 'XOR' operator is because of its properties. 

• Properties of 'XOR' operator. 
  
  • 1^1 = 0
  • 0^0 = 0
  • 1^0 = 1
  • 0^1 = 1
• If two bits are different then the 'XOR' operator returns a set bit(1) else it returns an unset bit(0).

6

Time Complexity: O(1)
Auxiliary Space: O(1)

4. Checking if bit at nth position is set or unset: 

It is quite easily doable using the 'AND' operator.

• Left shift '1' to given position and then 'AND'('&').

1

Time Complexity: O(1)
Auxiliary Space: O(1)

Observe that we have first left shifted '1' and then used 'AND' operator to get bit at that position. So if there is '1' at position 'pos' in 'num', then after 'AND' our variable 'bit' will store '1' else if there is '0' at position 'pos' in the number 'num' than after 'AND' our variable bit will store '0'.

Some more quick hacks: 

• Inverting every bit of a number/1's complement: If we want to invert every bit of a number i.e change bit '0' to '1' and bit '1' to '0'.We can do this with the help of '~' operator. For example : if number is num=00101100 (binary representation) so '~num' will be '11010011'.

This is also the '1s complement of number'.

-5

Time Complexity: O(1)
Auxiliary Space: O(1)

• Two's complement of the number: 2's complement of a number is 1's complement + 1.

So formally we can have 2's complement by finding 1s complement and adding 1 to the result i.e (~num+1) or what else we can do is using '-' operator.

This is two's complement -4
This is also two's complement -4

Time Complexity: O(1)
Auxiliary Space: O(1)

 Stripping off the lowest set bit :

In many situations we want to strip off the lowest set bit for example in Binary Indexed tree data structure, counting number of set bit in a number. We do something like this:  

X = X & (X-1)

But how does it even work? Let us see this by taking an example, let X = 1100.

• (X-1)  inverts all the bits till it encounters the lowest set '1' and it also inverts that lowest set '1'.
• X-1 becomes 1011. After 'ANDing' X with X-1 we get the lowest set bit stripped. 

6

Time Complexity: O(1)
Auxiliary Space: O(1)

Getting the lowest set bit of a number:

This is done by using the expression 'X &(-X)'Let us see this by taking an example: Let X = 00101100. So ~X(1's complement) will be '11010011' and 2's complement will be (~X+1 or -X) i.e.  '11010100'.So if we 'AND' original number 'X' with its two's complement which is '-X', we get the lowest set bit. 

 00101100
& 11010100
-----------
 00000100

2

Time Complexity: O(1)
Auxiliary Space: O(1)

Division by 2 and Multiplication by 2 are very frequently that too in loops in Competitive Programming so using Bitwise operators can help in speeding up the code.

Divide by 2 using the right shift operator:

00001100 >> 1 (00001100 is 12)
------------
00000110 (00000110 is 6)

6

Time Complexity: O(1)
Auxiliary Space: O(1)

Multiply by 2 using the left shift operator:

00001100 << 1 (00001100 is 12)
------------
00011000 (00000110 is 24)

24

Time Complexity: O(1)
Auxiliary Space: O(1)

Bit Tricks for Competitive Programming
Refer BitWise Operators Articles for more articles on Bit Hacks.