1534. Count Good Triplets
EasyLeetCodeGiven 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.
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