Introduction

In this blog post, we will explore how to solve the LeetCode problem "477" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. Example 1: 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. Example 2: Input: nums = [4,14,4] Output: 4 Constraints: 1 <= nums.length <= 104 0 <= nums[i] <= 109 The answer for the given input will fit in a 32-bit integer.

Explanation

Here's an efficient solution to calculate the sum of Hamming distances between all pairs of integers in an array:

• Bitwise Analysis: Instead of calculating the Hamming distance between each pair directly, analyze each bit position (0 to 31) across all numbers.
• Count 1s and 0s: For each bit position, count how many numbers have a '1' and how many have a '0' at that position.

• Calculate Contribution: The Hamming distance contribution for that bit position is simply the product of the count of 1s and the count of 0s. Sum this contribution across all bit positions.

• Time Complexity: O(N), where N is the number of elements in nums. Space Complexity: O(1).

Code

    def totalHammingDistance(nums):
    """
    Calculates the sum of Hamming distances between all pairs of integers in nums.

    Args:
        nums: A list of integers.

    Returns:
        The sum of Hamming distances between all pairs in nums.
    """
    total_distance = 0
    n = len(nums)

    for i in range(32):  # Iterate through each bit position (0 to 31)
        count_1 = 0
        for num in nums:
            if (num >> i) & 1:  # Check if the i-th bit is set (1)
                count_1 += 1
        count_0 = n - count_1  # Number of elements with 0 at the i-th bit
        total_distance += count_1 * count_0  # Add contribution of this bit

    return total_distance