Easy XOR + Count Bit Manipulation
In plain English: Count how many bits are different between two numbers.
XOR produces 1 at every position where the bits differ — then count the set bits in the XOR result.
Prompt
Find the number of positions at which the corresponding bits of two integers differ.
Try to write it from scratch before scrolling down.
Solution
def hamming_distance(x, y):
xor = x ^ y # 1s mark where bits differ
count = 0
while xor:
xor &= xor - 1 # clear lowest set bit
count += 1
return count
# Test: hamming_distance(1, 4) == 2 (001 vs 100)
# Test: hamming_distance(3, 1) == 1 (11 vs 01)