# Solving Leetcode Interviews in Seconds with AI: Count Good Triplets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1534" 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
	> Given an array of integers arr, and three integers a, b and c. You need to find the number of good triplets. A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:  0 <= i < j < k < arr.length |arr[i] - arr[j]| <= a |arr[j] - arr[k]| <= b |arr[i] - arr[k]| <= c  Where |x| denotes the absolute value of x. Return the number of good triplets.   Example 1: 
 Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
 Output: 4
 Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].
  Example 2: 
 Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1
 Output: 0
 Explanation: No triplet satisfies all conditions.
    Constraints:  3 <= arr.length <= 100 0 <= arr[i] <= 1000 0 <= a, b, c <= 1000 

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate and Check:** The most straightforward approach is to iterate through all possible triplets in the array and check if each triplet satisfies the given conditions.
*   **Direct Condition Evaluation:** For each triplet, we directly evaluate the absolute difference conditions `|arr[i] - arr[j]| <= a`, `|arr[j] - arr[k]| <= b`, and `|arr[i] - arr[k]| <= c`.
*   **Count Good Triplets:** If all three conditions are met, we increment a counter representing the number of good triplets.

*   **Runtime Complexity: O(n^3), Storage Complexity: O(1)**

	
	# Code
	```python
	def countGoodTriplets(arr, a, b, c):
    """
    Counts the number of "good" triplets in an array according to the given conditions.

    Args:
        arr: A list of integers.
        a: The maximum allowed absolute difference between arr[i] and arr[j].
        b: The maximum allowed absolute difference between arr[j] and arr[k].
        c: The maximum allowed absolute difference between arr[i] and arr[k].

    Returns:
        The number of good triplets in the array.
    """

    n = len(arr)
    count = 0

    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if abs(arr[i] - arr[j]) <= a and abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
                    count += 1

    return count
	```