# Solving Leetcode Interviews in Seconds with AI: Number of Arithmetic Triplets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2367" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:  i < j < k, nums[j] - nums[i] == diff, and nums[k] - nums[j] == diff.  Return the number of unique arithmetic triplets.   Example 1:  Input: nums = [0,1,4,6,7,10], diff = 3 Output: 2 Explanation: (1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3. (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.   Example 2:  Input: nums = [4,5,6,7,8,9], diff = 2 Output: 2 Explanation: (0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2. (1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.    Constraints:  3 <= nums.length <= 200 0 <= nums[i] <= 200 1 <= diff <= 50 nums is strictly increasing.  

	# Explanation
	Here's a breakdown of the solution:

*   **Hashing for Efficient Lookup:** Utilize a set (or hash table) to store all numbers in `nums`. This enables O(1) average time complexity for checking if a number exists in the array.
*   **Iterate and Check:** Iterate through the `nums` array. For each number, check if `num + diff` and `num + 2*diff` are present in the set. If both exist, increment the count of arithmetic triplets.
*   **Uniqueness:** Since `nums` is strictly increasing, the triplets found will automatically be unique, satisfying the `i < j < k` condition.

*   **Runtime Complexity:** O(n), where n is the length of `nums`.
*   **Storage Complexity:** O(n), where n is the length of `nums` (due to the set).

	
	# Code
	```python
	def arithmeticTriplets(nums: list[int], diff: int) -> int:
    """
    Finds the number of unique arithmetic triplets in a strictly increasing array.

    Args:
        nums: A 0-indexed, strictly increasing integer array.
        diff: A positive integer representing the difference between elements in the triplet.

    Returns:
        The number of unique arithmetic triplets.
    """
    count = 0
    num_set = set(nums)

    for num in nums:
        if (num + diff) in num_set and (num + 2 * diff) in num_set:
            count += 1

    return count
	```