Leetcode #477: Total Hamming Distance
In this guide, we solve Leetcode #477 Total Hamming Distance in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given an integer array nums, return the sum of Hamming distances between all the pairs of the integers in nums.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, Math
Intuition
The problem structure lets us track state with bitwise operations.
Bit operations are constant time and avoid extra memory.
Approach
Apply XOR/AND/OR and shifts to maintain the required invariant.
Aggregate the result in a single pass.
Steps:
- Identify a bitwise invariant.
- Combine values with bit operations.
- Return the aggregated result.
Example
Input: nums = [4,14,2]
Output: 6
Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case).
The answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
Python Solution
class Solution:
def totalHammingDistance(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for i in range(32):
a = sum(x >> i & 1 for x in nums)
b = n - a
ans += a * b
return ans
Complexity
The time complexity is , where and are the length of the array and the maximum value in the array, respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.