[Leetcode] 1534. Count Good Triplets

initial

this is just very basic triple for loop

sol

but need to care the "continue" step cuz we dont wanna early break.

class Solution:
    def countGoodTriplets(self, arr: List[int], a: int, b: int, c: int) -> int:
        length = len(arr)
        ans=0
        for i in range(length-2):
            for j in range(i+1,length-1):
                if abs(arr[i]-arr[j])>a:
                    continue
                for k in range(j+1,length):
                    if abs(arr[j]-arr[k])<=b and abs(arr[i]-arr[k])<=c:
                        ans+=1
        return ans

complexity

n^3 time
1 space