Introduction

In this blog post, we will explore how to solve the LeetCode problem "2006" 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 integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k. The value of |x| is defined as: x if x >= 0. -x if x < 0. Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Explanation: The pairs with an absolute difference of 1 are: - [1,2,2,1] - [1,2,2,1] - [1,2,2,1] - [1,2,2,1] Example 2: Input: nums = [1,3], k = 3 Output: 0 Explanation: There are no pairs with an absolute difference of 3. Example 3: Input: nums = [3,2,1,5,4], k = 2 Output: 3 Explanation: The pairs with an absolute difference of 2 are: - [3,2,1,5,4] - [3,2,1,5,4] - [3,2,1,5,4] Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100 1 <= k <= 99

Explanation

• The core idea is to iterate through the array and, for each element nums[i], count the number of elements nums[j] (where j > i) such that the absolute difference between nums[i] and nums[j] is equal to k.
  • To optimize this, we can use a hash map (or in this case, since the constraints limit values from 1-100, a simple array/list) to store the frequency of each number encountered so far. This lets us quickly check how many numbers exist such that nums[j] = nums[i] + k or nums[j] = nums[i] - k.
  • Iterate through the array. For each element, instead of comparing to the rest of the array, we will use the frequency array to determine the number of valid pairs.

• Runtime Complexity: O(n), where n is the length of the input array nums. Storage Complexity: O(1), because the frequency array's size is bounded by the constant range of possible values (1-100 inclusive).

Code

    def countKDifference(nums: list[int], k: int) -> int:
    """
    Given an integer array nums and an integer k, return the number of pairs (i, j)
    where i < j such that |nums[i] - nums[j]| == k.
    """
    count = 0
    freq = [0] * 101  # Initialize frequency array for numbers from 1 to 100

    for num in nums:
        count += (freq[num + k] if num + k <= 100 else 0) + (freq[num - k] if num - k >= 1 else 0)
        freq[num] += 1

    return count