# LeetCode 1534. Count Good Triplets

## Description

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

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

Just implement as the problem description is.

## Python Solution

``````class Solution:
def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
results = []

for i in range(0, len(arr)):
for j in range(i + 1, len(arr)):
for k in range(j + 1, len(arr)):
if abs(arr[i] - arr[j]) <= a and abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
results.append([i, j, k])

return len(results)``````
• Time Complexity: O(N^3)
• Space Complexity: O(N)