LeetCode 2367. Number of Arithmetic Triplets

Question

You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:

• i < j < k,
• nums[j] - nums[i] == diff, and
• nums[k] - nums[j] == diff.

Return the number of unique arithmetic triplets.

 

Example 1:

Input: nums = [0,1,4,6,7,10], diff = 3
Output: 2
Explanation:
(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.
(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3. 

 

Example 2:

Input: nums = [4,5,6,7,8,9], diff = 2
Output: 2
Explanation:
(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.
(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.

 

Constraints:

• 3 <= nums.length <= 200
• 0 <= nums[i] <= 200
• 1 <= diff <= 50
• nums is strictly increasing.

 

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Solution

Python

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        count = 0
        for i in range(len(nums)-2):
            for j in range(i,len(nums)-1):
                for k in range(j,len(nums)):
                    if nums[j] - nums[i] == diff and nums[k] - nums[j] == diff:
                        count += 1
    
        return count

Java