How do you solve Number of Arithmetic Triplets on LeetCode?

Number of Arithmetic Triplets (LeetCode #2367) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashSet allows for O(1) average time complexity for lookups.

What is the brute force solution for Number of Arithmetic Triplets?

The brute force approach for Number of Arithmetic Triplets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible triplet in the array to see if they meet the conditions for being an arithmetic triplet. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Arithmetic Triplets?

Number of Arithmetic Triplets uses the following concepts: Array, Hash Table, Two Pointers, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Arithmetic Triplets?

Always consider edge cases, like the minimum length of the array. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice identifying patterns in problems to recognize when to use specific data structures.

What are common mistakes when solving Number of Arithmetic Triplets?

Not using a set for quick lookups, leading to unnecessary nested loops. Overlooking the strictly increasing property of the array which can simplify checks.

#2367

Number of Arithmetic Triplets

Easy

ArrayHash TableTwo PointersEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using a HashSet allows us to quickly check if the required numbers to form an arithmetic triplet exist, reducing the need for nested loops and improving efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a HashSet to store the numbers in the array for quick lookup.
2Step 2: Iterate through the array and for each number, check if both (num + diff) and (num + 2 * diff) exist in the HashSet.
3Step 3: If both numbers exist, increment the count of valid triplets.

solution.py12 lines

1# Full working Python code
2
3def count_arithmetic_triplets(nums, diff):
4    num_set = set(nums)
5    count = 0
6    for num in nums:
7        if (num + diff) in num_set and (num + 2 * diff) in num_set:
8            count += 1
9    return count
10
11# Example usage
12print(count_arithmetic_triplets([0,1,4,6,7,10], 3))

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a couple of times. The space complexity is O(n) due to the HashSet storing the numbers for quick lookup.

1Using a HashSet allows for O(1) average time complexity for lookups.
2The problem can be solved efficiently by recognizing the arithmetic sequence properties.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.