Introduction

In this blog post, we will explore how to solve the LeetCode problem "532" 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

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true: 0 <= i, j < nums.length i != j |nums[i] - nums[j]| == k Notice that |val| denotes the absolute value of val. Example 1: Input: nums = [3,1,4,1,5], k = 2 Output: 2 Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5). Although we have two 1s in the input, we should only return the number of unique pairs. Example 2: Input: nums = [1,2,3,4,5], k = 1 Output: 4 Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5). Example 3: Input: nums = [1,3,1,5,4], k = 0 Output: 1 Explanation: There is one 0-diff pair in the array, (1, 1). Constraints: 1 <= nums.length <= 104 -107 <= nums[i] <= 107 0 <= k <= 107

Explanation

Here's an efficient solution to find the number of unique k-diff pairs in an array:

• Hashing for Efficiency: Use a hash map (dictionary in Python) to store the frequency of each number in the input array. This allows for O(1) lookups when searching for pairs.
• Handle k = 0 separately: If k is 0, we only need to count the numbers that appear more than once.

• Avoid Duplicates: Use a set to keep track of the unique pairs that have already been counted to avoid overcounting.

• Time Complexity: O(n), where n is the length of the input array.

• Space Complexity: O(n), where n is the length of the input array (for the hash map and the set).

Code

    def k_diff_pairs(nums, k):
    """
    Finds the number of unique k-diff pairs in an array.

    Args:
        nums: A list of integers.
        k: An integer representing the difference.

    Returns:
        The number of unique k-diff pairs in the array.
    """

    count = 0
    seen_pairs = set()
    num_counts = {}

    for num in nums:
        num_counts[num] = num_counts.get(num, 0) + 1

    if k == 0:
        for num, freq in num_counts.items():
            if freq > 1:
                count += 1
    else:
        for num in num_counts:
            if num + k in num_counts:
                pair = tuple(sorted((num, num + k)))
                if pair not in seen_pairs:
                    count += 1
                    seen_pairs.add(pair)

    return count