1534. Count Good Triplets

EasyLeetCode

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

How to solve the problem

• Adding the first if statement if abs(arr[i] - arr[j]) ≤ a: before the innermost loop, we can skip the entire innermost loop if the first condition is not satisfied.

• Using and for logic connection instead of bitwise AND & in Python.

• If a method is in class, we need the first parameter self to call the instance of class. And also we need an instance to call the method. Otherwise, using @staticmethod outside class.

Code

class Solution:
    def countGoodTriplets(self, arr: list[int], a: int, b: int, c: int) -> int:
        count = 0
        for i in range(len(arr)):
                for j in range(i + 1 ,len(arr)):
                    if abs(arr[i] - arr[j]) <= a:
                        for k in range(j + 1, len(arr)):
                            if abs(arr[j] - arr[k]) <= b and abs(arr[i] - arr[k]) <= c:
                                count += 1
        return count

Complexity

Time complexity: O(n³)

We're using 3 nested loops to go through every possible triplet.

Space complexity: O(1)

Only using one counter for the result.