Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

The input is small enough to examine every ordered index triple directly. Each good triple must satisfy all three pairwise distance limits; satisfying two does not automatically guarantee the third.

Approach

Use three nested loops with i < j < k. Check abs(arr[i] - arr[j]) <= a, abs(arr[j] - arr[k]) <= b, and abs(arr[i] - arr[k]) <= c, incrementing the answer only when all hold. Return the count. Increasing index bounds prevent duplicate counting while still distinguishing equal-valued elements at different positions. Equality with a threshold is allowed.

class Solution:
    def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
        n = len(arr)
        res = 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:
                        res += 1

        return res

Complexity analysis

For n = arr.length, the code checks every index triple, so time is O(n³) and auxiliary space is O(1).