Bit Tricks for Competitive Programming
In competitive programming or in general, some problems seem difficult but can be solved very easily with little concepts of bit magic. We have discussed some tricks below in the previous post.
Bitwise Hacks for Competitive Programming
OneLiner Hacks of Bit Manipulation:
OneLiner Code 
Function 







x & ~((1 << i+1 ) – 1) 
Clears all bits of x from LSB to ith bit 
x & ((1 << i) – 1) 
Clears all bits of x from MSB to ith bit 
x >>1 
Divides x by 2 
x << 1 
Multiplies x by 2 
ch  ‘ ‘ 
Upper case English alphabet ch to lower case 
ch & ‘_â€™ 
Lower case English alphabet ch to upper case 
x && !(x & x1) 
Checking if given 32bit integer is power of 2 
log2(n & n)+1 
Find the last set bit 
1) Check Whether Number is Even or Odd
x & 1
Logic:
Even numbers have 0 as their LSB and Odd numbers have 1 as their LSB, so Bitwise AND with even numbers results in 0 and with odd numbers results in 1.
Example:
x = 29 (00011101), (x & 1) = 1, therefore x is an odd number
x= 16(10000), (x & 1) = 0, therefore x is an even number
2) Clear the Lowest Set bit of x
x & (x1)
Logic:
When you subtract 1 from x, all the bits to the right of the lowest set bit become 1, and the lowest set bit becomes 0 and performing a bitwise AND operation between x and (x1) sets each bit to 0 except for the bits that are common in both x and (x1).
Example:
x = 10101010
x1 = 10101001
x & (x1) = 10101000
3) Extract the Lowest Set bit of x
x & ~(x1);
Logic:
When you subtract 1 from x, all the bits to the right of the lowest set bit become 1, and the lowest set bit becomes 0 and performing a bitwise AND operation between x and ~(x1) sets each bit to 0 except for the lowest set bit of x.
Example:
x = 10101010
~(x1) = 01010110
x & ~(x1) = 00000010
4) Clear all bits from LSB to ith bit
mask = ~((1 << i+1 )  1);
x &= mask;
Logic:
To clear all bits from LSB to ith bit, we have to AND x with mask having LSB to ith bit 0. To obtain such mask, first left shift 1 i times. Now if we minus 1 from that, all the bits from 0 to i1 become 1 and remaining bits become 0. Now we can simply take the complement of mask to get all first i bits to 0 and remaining to 1.
Example:–
x = 29 (00011101) and we want to clear LSB to 3rd bit, total 4 bits
mask > 1 << 4 > 16(00010000)
mask > 16 â€“ 1 > 15(00001111)
mask > ~mask > 11110000
x & mask > 16 (00010000)
5) Clearing all bits from MSB to ith bit
mask = (1 << i)  1;
x &= mask;
Logic:
To clear all bits from MSB to ith bit, we have to AND x with mask having MSB to ith bit 0. To obtain such mask, first left shift 1 i times. Now if we minus 1 from that, all the bits from 0 to i1 become 1 and the remaining bits become 0.
Example:
x = 215 (11010111) and we want to clear MSB to 4th bit, total 4 bits
mask > 1 << 4 > 16(00010000)
mask > 16 â€“ 1 > 15(00001111)
x & mask > 7(00000111)
6) Divide by 2
x >>= 1;
Logic:
When we do arithmetic right shift, every bit is shifted to right and blank position is substituted with sign bit of number, 0 in case of positive and 1 in case of negative number. Since every bit is a power of 2, with each shift we are reducing the value of each bit by factor of 2 which is equivalent to division of x by 2.
Example:
x = 18(00010010)
x >> 1 = 9 (00001001)
7) Multiplying by 2
x <<= 1;
Logic:
When we do arithmetic left shift, every bit is shifted to left and blank position is substituted with 0 . Since every bit is a power of 2, with each shift we are increasing the value of each bit by a factor of 2 which is equivalent to multiplication of x by 2.
Example:
x = 18(00010010)
x << 1 = 36 (00100100)
8) Upper case English alphabet to lower case
ch = ' ';
Logic: The bit representation of upper case and lower case English alphabets are –
A > 01000001 a > 01100001
B > 01000010 b > 01100010
C > 01000011 c > 01100011
. .
. .
Z > 01011010 z > 01111010
As we can see if we set 5th bit of upper case characters, it will be converted into lower case character. We have to prepare a mask having 5th bit 1 and other 0 (00100000). This mask is bit representation of space character (â€˜ â€˜). The character â€˜châ€™ then OR with mask.
Example:
ch = â€˜Aâ€™ (01000001)
mask = â€˜ â€˜ (00100000)
ch  mask = â€˜aâ€™ (01100001)
Please refer Case conversion (Lower to Upper and Vice Versa) for details.
9) Lower case English alphabet to upper case
ch &= '_â€™ ;
Logic: The bit representation of upper case and lower case English alphabets are –
A > 01000001 a > 01100001
B > 01000010 b > 01100010
C > 01000011 c > 01100011
. .
. .
Z > 01011010 z > 01111010
As we can see if we clear 5th bit of lowercase characters, it will be converted into upper case character. We have to prepare a mask having 5th bit 0 and other 1 (11011111). This mask is bit representation of underscore character (â€˜_â€˜). The character â€˜châ€™ then AND with mask.
Example:
ch = â€˜aâ€™ (01100001)
mask = â€˜_ â€˜ (11011111)
ch & mask = â€˜Aâ€™ (01000001)
Please refer Case conversion (Lower to Upper and Vice Versa) for details.
10) Count set bits in integer
int countSetBits(int x)
{
int count = 0;
while (x)
{
x &= (x1);
count++;
}
return count;
}
Logic: This is Brian Kernighanâ€™s algorithm.
11) Find log base 2 of 32 bit integer
int log2(int x)
{
int res = 0;
while (x >>= 1)
res++;
return res;
}
Logic: We right shift x repeatedly until it becomes 0, meanwhile we keep count on the shift operation. This count value is the log2(x).
12) Checking if given 32bit integer is power of 2
int isPowerof2(int x)
{
return (x && !(x & x1));
}
Logic:
All the power of 2 have only single bit set e.g. 16 (00010000). If we minus 1 from this, all the bits from LSB to set bit get toggled, i.e., 161 = 15 (00001111). Now if we AND x with (x1) and the result is 0 then we can say that x is power of 2 otherwise not. We have to take extra care when x = 0.
Example :
x = 16(00010000)
x â€“ 1 = 15(00001111)
x & (x1) = 0
so, 16 is power of 2
13) Find the last set bit
int lastSetBit(int n){
return log2(n & n)+1;
}
The logarithmic value of AND of x and x to the base 2 gives the index of the last set bit(for 0based indexing).
Playing with Kth bit:
1) Turn Off k^{th} bit in a number
int turnOffKthBit(int n, int k) {
return n & ~(1 << (k  1));
}
2) Turn On k^{th} bit in a number
int turnOnKthBit(int n, int k) {
return n  (1 << (k  1));
}
3) Check if k^{th} bit is set for a number
bool isKthBitSet(int n, int k) {
return (n & (1 << (k  1))) != 0;
}
4) Toggle the k^{th} bit
int toggleKthBit(int n, int k) {
return n ^ (1 << (k  1));
}
Please refer this article for more bit hacks.
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