# Solving Leetcode Interviews in Seconds with AI: Count Subarrays of Length Three With a Condition


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3392" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate through subarrays of length 3:** The core idea is to examine each possible subarray of length 3 within the given array.
*   **Check the condition:** For each subarray, verify if the sum of the first and third elements is equal to half of the second element.
*   **Count valid subarrays:** Increment a counter each time the condition is met.

*   **Runtime Complexity:** O(n), where n is the length of the input array. **Storage Complexity:** O(1).

	
	# Code
	```python
	def count_subarrays(nums):
    """
    Counts the number of subarrays of length 3 where the sum of the first and third
    numbers equals half of the second number.

    Args:
        nums: A list of integers.

    Returns:
        The number of subarrays that satisfy the condition.
    """
    count = 0
    for i in range(len(nums) - 2):
        if nums[i] + nums[i + 2] == nums[i + 1] / 2:
            count += 1
    return count
	```