Understanding the Problem
We are given two integers a
and b
. Our task is to find out how many bits are different between their binary representations. In other words, we need to count how many bits we need to flip in a
so that it becomes equal to b
.
Each bit can be either 0 or 1. A bit flip means changing a 0 to a 1, or a 1 to a 0. This problem is a great exercise in using bitwise operations, especially XOR
, to compare bits.
Step-by-Step Solution with Example
Step 1: Understand XOR (Exclusive OR)
XOR is a bitwise operation that compares each corresponding bit of two numbers. It returns 1 if the bits are different and 0 if they are the same.
For example:
a = 10 -> binary: 1010
b = 20 -> binary: 10100
To make bit lengths equal:
a = 01010
b = 10100
XOR:
01010
^ 10100
--------
11110
The result is 11110
in binary, which is 30
in decimal. This result has 4 bits set to 1, which means we need to flip 4 bits in a
to get b
.
Step 2: Initialize a counter
We’ll use a counter variable, initially set to 0, to keep track of how many 1’s are in the XOR result.
Step 3: Loop through each bit of the XOR result
We check the last bit of the XOR result using xor & 1
. If it’s 1, we increment the counter. Then we right shift the number using xor >>= 1
to move to the next bit. We continue until the XOR result becomes 0.
Step 4: Return the counter
The counter now holds the number of differing bits between a
and b
.
Edge Cases
- When a and b are equal: XOR result will be 0. Hence, 0 bits need to be flipped.
- When either a or b is 0: The number of flips is equal to the number of 1's in the other number's binary representation.
- Very large numbers: The algorithm still works efficiently as it only processes the number of bits set in the XOR result.
Final Thoughts
This problem is a classic use of the XOR operation to compare bit patterns. It helps build intuition around binary representation and how bitwise operations work. It’s also efficient, as it doesn’t need to convert numbers to strings or use arrays — everything is done using bitwise math.
Understanding how binary numbers differ and using simple looping allows us to solve this problem in a clean and optimized way, which is essential for both interviews and real-world programming.
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