Count Subarrays of Length Three With a Condition Python/Java solution

In both Python and Java, iterate through the array from index 0 to n-3 and check whether nums[i] + nums[i+2] == 2 * nums[i+1]. Increment a counter whenever the condition holds. The implementation is short and runs in O(n) time with O(1) space.

How to solve Count Subarrays of Length Three With a Condition in O(n)?

Traverse the array once and evaluate every contiguous triple (nums[i], nums[i+1], nums[i+2]). For each triple, check the condition nums[i] + nums[i+2] == 2 * nums[i+1]. If it holds, increment a counter. This single pass guarantees O(n) time and constant memory usage.

What is the best approach for Count Subarrays of Length Three With a Condition?

The best approach is a single pass sliding window over the array. Iterate from index 0 to n-3 and check whether nums[i] + nums[i+2] equals 2 * nums[i+1]. This verifies that the middle element is the average of its neighbors. The algorithm runs in O(n) time with O(1) extra space.

Is Count Subarrays of Length Three With a Condition asked at Google/Amazon/Meta?

Problems of this style appear frequently in interviews at companies like Amazon, Google, and Meta because they test array traversal and simple mathematical reasoning. While this exact problem may come from LeetCode contests, the pattern of scanning fixed-size windows is very common in interviews.

What data structure is used in Count Subarrays of Length Three With a Condition?

The problem primarily uses a simple array traversal. No advanced data structures are required. A fixed-size sliding window of three elements and basic arithmetic checks are enough to implement the optimal solution.

What is the time complexity of Count Subarrays of Length Three With a Condition?

The optimal solution runs in O(n) time because you examine each possible length‑3 window exactly once. There are n-2 such windows in an array of size n. Each check is constant time, and the algorithm uses O(1) additional space.

Description Hints 1 Solution Video 6:55

3392. Count Subarrays of Length Three With a Condition

Q: Is Count Subarrays of Length Three With a Condition easy or hard?

LeetCode classifies this problem as Easy. The logic only requires scanning the array once and checking a simple arithmetic relationship among three consecutive elements. It mainly tests familiarity with array iteration and recognizing patterns like arithmetic progressions.

Easy61.5% acceptance5 companies

Given an integer array nums, return the number of subarrays of length 3 such that the sum of the first and third numbers equals exactly half of the second number.

Example 1:

Input: nums = [1,2,1,4,1]

Output: 1

Explanation:

Only the subarray [1,4,1] contains exactly 3 elements where the sum of the first and third numbers equals half the middle number.

Example 2:

Input: nums = [1,1,1]

Output: 0

Explanation:

[1,1,1] is the only subarray of length 3. However, its first and third numbers do not add to half the middle number.

Constraints:

3 <= nums.length <= 100
-100 <= nums[i] <= 100

Asked at 5 companies

Cognizant Amazon

Google Microsoft Meta

Topics

Array

Hints open one at a time

Hint 1

The constraints are small. Consider checking every subarray.

6:55

Still stuck?Read the written solution, or watch a walkthrough.

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1class Solution:
2    def countSubarrays(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,2,1,4,1]

Input

[1,1,1]

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