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.
C++
#include<iostream>
using namespace std;
void set(int & num,int pos)
{
num |= (1 << pos);
}
int main()
{
int num = 4, pos = 1;
set(num, pos);
cout << (int)(num) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int num = 4, pos =1;
num = set(num, pos);
System.out.println(num);
}
public static int set(int num, int pos){
num |= (1 << pos);
return num;
}
}
|
Python3
def set (num, pos):
num |= (1 << pos)
print(num)
num, pos = 4, 1
set(num, pos)
|
C#
using System;
public class GFG{
static public void Main ()
{
int num = 4, pos = 1;
set(num, pos);
}
static public void set(int num, int pos)
{
num |= (1 << pos);
Console.WriteLine(num);
}
}
|
Javascript
<script>
function set(num,pos)
{
num |= (1 << pos);
console.log(parseInt(num));
}
let num = 4;
let pos = 1;
set(num, pos);
</script>
|
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.
C++
#include <iostream>
using namespace std;
void unset(int &num,int pos)
{
num &= (~(1 << pos));
}
int main()
{
int num = 7;
int pos = 1;
unset(num, pos);
cout << num << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 7, pos = 1;
num = unset(num, pos);
System.out.println(num);
}
public static int unset(int num, int pos)
{
num = num & (~(1 << pos));
return num;
}
}
|
Python3
def unset(num, pos):
num &= (~(1 << pos))
print(num)
num, pos = 7, 1
unset(num, pos)
|
C#
using System;
public class GFG {
static public void Main()
{
int num = 7, pos = 1;
unset(num, pos);
}
static public void unset(int num, int pos)
{
num &= (~(1 << pos));
Console.WriteLine(num);
}
}
|
Javascript
function unset(num,pos)
{
num &= ( ~ (1 << pos));
return num;
}
let num = 7;
let pos = 1;
console.log(unset(num, pos));
|
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).
C++
#include <iostream>
using namespace std;
void toggle(int& num, int pos) { num ^= (1 << pos); }
int main()
{
int num = 4;
int pos = 1;
toggle(num, pos);
cout << num << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int num = 4, pos =1;
num = toggle(num, pos);
System.out.println(num);
}
public static int toggle(int num, int pos){
num ^= (1 << pos);
return num;
}
}
|
Python3
def toggle(num, pos):
num ^= (1 << pos)
print(num)
num, pos = 4, 1
toggle(num, pos)
|
C#
using System;
public class GFG{
static public void Main ()
{
int num = 4, pos = 1;
toggle(num, pos);
}
static public void toggle(int num, int pos)
{
num ^= (1 << pos);
Console.WriteLine(num);
}
}
|
Javascript
function toggle(num, pos) {
num ^= (1 << pos);
return num;
}
let num = 4,
pos = 1;
num = toggle(num, pos);
console.log(num);
|
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'(‘&’).
C++
#include <iostream>
using namespace std;
bool at_position(int num, int pos)
{
bool bit = num & (1 << pos);
return bit;
}
int main()
{
int num = 5;
int pos = 0;
bool bit = at_position(num, pos);
cout << bit << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 5;
int pos = 0;
int bit = at_position(num, pos);
System.out.println(bit);
}
public static int at_position(int num, int pos)
{
int bit = num & (1 << pos);
return bit;
}
}
|
Python3
def at_position(num,pos):
bit = num & (1<<pos)
return bit
num = 5
pos = 0
bit = at_position(num, pos)
print(bit)
|
C#
using System;
class GFG {
static void Main(string[] args)
{
int num = 5;
int pos = 0;
int bit = at_position(num, pos);
Console.WriteLine(bit);
}
static int at_position(int num, int pos)
{
int bit = num & (1 << pos);
return bit;
}
}
|
Javascript
<script>
function at_position(num,pos)
{
return num & (1<<pos);
}
let num = 5;
let pos = 0;
console.log(at_position(num, pos));
</script>
|
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’.
C++
#include <iostream>
using namespace std;
int main()
{
int num = 4;
cout << (~num);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int num = 4;
num = (~num);
System.out.println(num);
}
}
|
Python3
C#
using System;
public class GFG{
static public void Main ()
{
int num = 4;
Console.WriteLine(~num);
}
}
|
Javascript
<script>
let num = 4;
console.log(~num);
</script>
|
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.
C++
#include <iostream>
using namespace std;
int main()
{
int num = 4;
int twos_complement = -num;
cout << "This is two's complement " << twos_complement << endl;
cout << "This is also two's complement " << (~num+1) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 4;
int twos_complement = -num;
System.out.println("This is two's complement "
+ twos_complement);
System.out.println("This is also two's complement "
+ (~num + 1));
}
}
|
Python3
num = 4
twos_complement = -num
print(f"This is two's complement {twos_complement}")
print(f"This is also two's complement {~num + 1}")
|
C#
using System;
public class GFG{
static public void Main ()
{
int num = 4;
int twos_complement = -num;
Console.WriteLine("This is two's complement " + twos_complement);
Console.WriteLine("This is also two's complement " + (~num+1));
}
}
|
Javascript
<script>
let num = 4;
let twos_complement = -num;
console.log("This is two's complement " + twos_complement);
console.log("This is also two's complement " + (~num+1));
</script>
|
Output
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.
C++
#include <iostream>
using namespace std;
void strip_last_set_bit(int &num)
{
num = num & (num-1);
}
int main()
{
int num = 7;
strip_last_set_bit(num);
cout << num << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 7;
num = strip_last_set_bit(num);
System.out.println(num);
}
public static int strip_last_set_bit(int num)
{
return num & (num - 1);
}
}
|
Python3
def strip_last_set_bit(num):
num &= (num - 1)
print(num)
num = 7
strip_last_set_bit(num)
|
C#
using System;
public class GFG{
static public void Main ()
{
int num = 7;
strip_last_set_bit(num);
}
static public void strip_last_set_bit(int num)
{
num &= (num - 1);
Console.WriteLine(num);
}
}
|
Javascript
<script>
function strip_last_set_bit(num)
{
return num & (num-1);
}
let num = 7;
console.log(strip_last_set_bit(num));
</script>
|
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
C++
#include <iostream>
using namespace std;
int lowest_set_bit(int num)
{
int ret = num & (-num);
return ret;
}
int main()
{
int num = 10;
int ans = lowest_set_bit(num);
cout << ans << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 10;
int ans = lowest_set_bit(num);
System.out.println(ans);
}
public static int lowest_set_bit(int num)
{
int ret = num & (-num);
return ret;
}
}
|
Python3
def lowest_set_bit(num):
num &= (-num)
print(num)
num = 10
lowest_set_bit(num)
|
Javascript
function lowest_set_bit(num)
{
let ret = num & (-num);
return ret;
}
let num = 10
let ans = lowest_set_bit(num)
console.log(ans)
|
C#
using System;
public class GFG {
static public void Main()
{
int num = 10;
lowest_set_bit(num);
}
static public void lowest_set_bit(int num)
{
num &= (~num + 1);
Console.WriteLine(num);
}
}
|
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)
C++
#include <iostream>
using namespace std;
int main()
{
int num = 12;
int ans = num>>1;
cout << ans << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String[] args)
{
int num = 12;
int ans = num >> 1;
System.out.println(ans);
}
}
|
Python3
C#
using System;
public class GFG{
static public void Main ()
{
int num = 12;
Console.WriteLine(num >> 1);
}
}
|
Javascript
<script>
var num = 12;
var ans = num>>1;
console.log(ans);
</script>
|
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)
C++
#include <iostream>
using namespace std;
int main()
{
int num = 12;
int ans = num<<1;
cout << ans << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int num = 12;
int ans = num<<1;
System.out.println(ans);
}
}
|
C#
using System;
public class GFG{
static public void Main ()
{
int num = 12;
Console.WriteLine(num << 1);
}
}
|
Python3
num = 12
ans = num<<1
print(ans)
|
Javascript
<script>
var num = 12;
var ans = num<<1;
document.write(ans);
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Bit Tricks for Competitive Programming
Refer BitWise Operators Articles for more articles on Bit Hacks.
This article is contributed by Pankaj Mishra. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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