How do you solve Hamming Distance on LeetCode?

Hamming Distance (LeetCode #461) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: XOR operation is a powerful tool for bit manipulation.

What is the brute force solution for Hamming Distance?

The brute force approach for Hamming Distance is Brute Force, with O(32) or O(1) time complexity and O(1) space complexity. The brute force approach involves comparing each bit of the two integers to see if they differ. This method is straightforward but inefficient for larger integers. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Hamming Distance?

Hamming Distance uses the following concepts: Bit Manipulation. The recommended patterns to study are: Bit Manipulation, XOR.

What are the interview tips for Hamming Distance?

Explain your thought process clearly during the interview. Practice bit manipulation problems to become comfortable with concepts. Remember to consider both time and space complexity when discussing your solution.

What are common mistakes when solving Hamming Distance?

Forgetting to handle edge cases like both numbers being the same. Not understanding how to count bits efficiently.

#461

Hamming Distance

Easy

Bit ManipulationBit ManipulationXOR

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(32) or O(1)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

The optimal solution uses the XOR operation to find differing bits directly. By XORing the two numbers, we can identify all the differing bits in one go, making the solution more efficient.

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Algorithm

2 steps

1Step 1: Perform XOR between x and y to get a number that has bits set to 1 where x and y differ.
2Step 2: Count the number of 1s in the result of the XOR operation.

solution.py4 lines

1# Full working Python code
2
3def hamming_distance(x, y):
4    return bin(x ^ y).count('1')

ℹ

Complexity note: The time complexity is O(1) because the operations performed (XOR and counting bits) are constant time operations. The space complexity is O(1) as we are using a fixed amount of space.

1XOR operation is a powerful tool for bit manipulation.
2Counting bits can be efficiently done using built-in functions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.