[Leet Code] Count Good Triplets

Leetcode: https://leetcode.com/problems/count-good-triplets/

Problem:

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

Solution:

class Solution(object):
    def countGoodTriplets(self, arr, a, b, c):
        """
        :type arr: List[int]
        :type a: int
        :type b: int
        :type c: int
        :rtype: int
        """
        counter = 0
        
        for i in range(len(arr)):
            for j in range(len(arr)):
                for k in range(len(arr)):
                    if (abs(arr[i] - arr[j]) <= a) and (abs(arr[j] - arr[k]) <= b) and (abs(arr[i] - arr[k]) <= c) and (i < j) and (j < k):
                        counter += 1
        return counter