2367. Number of Arithmetic Triplets

제목

문제)

2367. Number of Arithmetic Triplets

 

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.

솔루션1)

- brute-force 방식이 가장 먼저 떠오름.

- 

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        """
        brute-force
        """
        count = 0
        for i, n in enumerate(nums):
            for j in range(i+1, len(nums)):
                if (nums[j] - nums[i]) == diff:
                    # found
                    for k in range(j+1, len(nums)):
                        if (nums[k] - nums[j]) == diff:
                            # found
                            count += 1                            
                            break
                if (nums[j] - nums[i]) > diff:
                    # not found
                    break
                    
        return count