#2367

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Number of Arithmetic Triplets

Count the number of arithmetic triplets in a given strictly increasing array with a specified difference.

array hash table two pointers enumeration

Problem Statement

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 hold: nums[j] - nums[i] == diff and nums[k] - nums[j] == diff. You need to return the number of unique arithmetic triplets.

Example 1: For nums = [0,1,4,6,7,10] and diff = 3, the valid triplets are (1, 2, 4) and (2, 4, 5), leading to the output 2. The task is to find a way to efficiently count these triplets for any given input.

Examples

Example 1

Input: nums = [0,1,4,6,7,10], diff = 3

Output: 2

(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

(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.

Solution Approach

Array Scanning with Hash Lookup

A direct approach to solving this problem is to iterate over each element in the array. For each number, check if both nums[i] + diff and nums[i] + 2*diff exist in the array. A hash set can be used to track seen elements, allowing constant time lookup to check if the required values exist.

Two Pointers Approach

After sorting the array, use a two-pointer technique to find pairs of elements with the given difference. For each pair, check if there exists a third element that forms an arithmetic triplet. This reduces unnecessary checks and improves efficiency.

Brute Force Enumeration

As a naive approach, you can use three nested loops to check all possible triplets within the array. This would involve iterating through the array and checking for every combination of triplet indices, but may be inefficient for large inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the chosen approach. The brute force method has O(n^3) complexity due to three nested loops. The hash lookup approach can reduce it to O(n) on average if hash operations are constant time. The two-pointer method can work in O(n log n) after sorting.

What Interviewers Usually Probe

The candidate understands the value of hash lookup for optimizing time complexity.
The candidate may propose multiple approaches with an understanding of trade-offs in performance.
The candidate mentions array scanning or two-pointer techniques to handle the problem efficiently.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle boundary conditions, such as ensuring the array is strictly increasing.
Overcomplicating the solution when a simple hash lookup or two-pointer technique would suffice.
Assuming that brute force is always an acceptable solution without considering time constraints for larger inputs.

Follow-up variants

What if the array was not strictly increasing? This could require a different approach to ensure the difference condition holds.
What if the diff value is much larger than the array size? This would require careful handling to avoid unnecessary checks.
Can the problem be adapted to find the number of arithmetic triplets in a non-zero-indexed array or with negative integers?

FAQ

How do I efficiently count arithmetic triplets in the given array?

Use a hash set to track the elements as you scan through the array. For each number, check if its 'diff' and '2*diff' counterparts exist in the set.

What is the time complexity of the brute force solution?

The brute force approach has a time complexity of O(n^3) due to three nested loops, which is inefficient for larger input sizes.

Can I solve this problem using only two-pointer technique?

Yes, by sorting the array first, you can apply the two-pointer technique to find valid triplets with the given difference.

What if the array is not strictly increasing?

If the array is not strictly increasing, you need to account for duplicates and might need to modify your approach accordingly.

How does GhostInterview help in solving this problem?

GhostInterview assists by providing structured approaches, identifying potential pitfalls, and offering insights into optimization strategies like hash lookup and two-pointer techniques.

terminal

Solution

Solution 1: Brute Force

We notice that the length of the array $nums$ is no more than $200$. Therefore, we can directly enumerate $i$, $j$, $k$, and check whether they meet the conditions. If they do, we increment the count of the triplet.

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2

3

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        return sum(b - a == diff and c - b == diff for a, b, c in combinations(nums, 3))

Solution 2: Array or Hash Table

We can first store the elements of $nums$ in a hash table or array $vis$. Then, for each element $x$ in $nums$, we check if $x+diff$ and $x+diff+diff$ are also in $vis$. If they are, we increment the count of the triplet.

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class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        return sum(b - a == diff and c - b == diff for a, b, c in combinations(nums, 3))

Continue Practicing

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