Is Number of Arithmetic Triplets easy or hard?

Number of Arithmetic Triplets is classified as an Easy problem on LeetCode with a high acceptance rate. The challenge mainly involves recognizing the arithmetic progression pattern and applying a hash set for efficient lookup.

How to solve Number of Arithmetic Triplets in O(n)?

Store all array values in a hash set. For each number x in nums, check if x + diff and x + 2*diff exist in the set. Every successful check forms one arithmetic triplet, so increment the count. The array is strictly increasing, which guarantees valid index ordering.

Number of Arithmetic Triplets Python or Java solution?

Most implementations use a set in Python or a HashSet in Java. Iterate through nums and check if num + diff and num + 2*diff exist in the set. The logic remains identical across Python, Java, C++, C#, and JavaScript.

What is the best approach for Number of Arithmetic Triplets?

The most efficient approach uses a hash set to check whether required elements exist. Insert all numbers into a set, then iterate through the array and verify if x + diff and x + 2*diff are present. This reduces the search to constant-time lookups and runs in O(n) time with O(n) space.

What data structure is used in Number of Arithmetic Triplets?

A hash set (hash table) is the key data structure. It enables O(1) average-time membership checks when verifying whether the next two numbers in the arithmetic progression exist.

What is the time complexity of Number of Arithmetic Triplets?

The optimal solution runs in O(n) time. Each element in the array is processed once, and two hash lookups are performed per element. Space complexity is O(n) due to storing the numbers in a hash set.

Is Number of Arithmetic Triplets asked at Google Amazon or Meta?

Arithmetic progression and hash set lookup problems commonly appear in interviews at companies like Amazon, Google, and Meta. Variations of this question test pattern recognition, hash table usage, and efficient array traversal.

Description Hints 2 Solution Video 4:46

2367. Number of Arithmetic Triplets

Easy85.4% acceptance2 companies

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.

Asked at 2 companies

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Hints open one at a time

Hint 1

Are the constraints small enough for brute force?

Hint 2

We can use three loops, each iterating through the array to go through every possible triplet. Be sure to not count duplicates.

4:46

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