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, 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

    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