Python Program to Check if nth Bit is Set

To check if the nth bit is set in Python, use the expression (num & (1 << n)) != 0, where num is the number and n is the bit position starting from 0.

Examples

Inputnum=5, n=0

OutputTrue

Inputnum=5, n=1

OutputFalse

Inputnum=8, n=3

OutputTrue

How to Think About It

To find if the nth bit is set, think of shifting the number 1 to the left by n positions to create a mask that has only the nth bit set. Then use the bitwise AND & operator between the number and this mask. If the result is not zero, the nth bit in the number is set; otherwise, it is not.

Algorithm

1

Get the input number and the bit position n.

2

Create a mask by shifting 1 left by n positions.

3

Perform bitwise AND between the number and the mask.

4

If the result is not zero, the nth bit is set; otherwise, it is not.

5

Return True if set, False if not.

Code

python

def is_nth_bit_set(num, n):
    return (num & (1 << n)) != 0

# Example usage
print(is_nth_bit_set(5, 0))  # True
print(is_nth_bit_set(5, 1))  # False
print(is_nth_bit_set(8, 3))  # True

Output

True
False
True

Dry Run

Let's trace the example num=5, n=1 through the code

1

Create mask

1 << 1 = 2 (binary 10)

2

Bitwise AND

5 & 2 = 0101 & 0010 = 0000 (decimal 0)

3

Check result

0 != 0 is False, so bit 1 is not set

Step	Operation	Value
1	Mask (1 << n)	2
2	num & mask	0
3	Result != 0	False

Why This Works

Step 1: Shift 1 to create mask

Shifting 1 left by n positions creates a number with only the nth bit set, which acts as a mask to isolate that bit.

Step 2: Use bitwise AND

The bitwise AND between the number and the mask keeps only the nth bit of the number; all other bits become zero.

Step 3: Check if bit is set

If the result of the AND is not zero, it means the nth bit was set in the original number.

Alternative Approaches

Using right shift and bitwise AND

python

def is_nth_bit_set(num, n):
    return ((num >> n) & 1) == 1

print(is_nth_bit_set(5, 0))  # True
print(is_nth_bit_set(5, 1))  # False
print(is_nth_bit_set(8, 3))  # True

This method shifts the number right by n bits and checks the least significant bit; it is equally efficient and sometimes easier to understand.

Using bin() string and indexing

python

def is_nth_bit_set(num, n):
    bin_str = bin(num)[2:][::-1]
    return n < len(bin_str) and bin_str[n] == '1'

print(is_nth_bit_set(5, 0))  # True
print(is_nth_bit_set(5, 1))  # False
print(is_nth_bit_set(8, 3))  # True

This method converts the number to a binary string and checks the character at position n; it is less efficient but useful for learning.

Complexity: O(1) time, O(1) space

Time Complexity

Checking a bit uses simple bitwise operations that run in constant time regardless of input size.

Space Complexity

Only a few variables are used, so space is constant.

Which Approach is Fastest?

Bitwise operations are fastest; string conversion is slower and less memory efficient.

Approach	Time	Space	Best For
Bitwise AND with mask	O(1)	O(1)	Fastest and simplest
Right shift and AND	O(1)	O(1)	Equally fast, easy to read
Binary string indexing	O(log n)	O(log n)	Learning and visualization