Leetcode 3392. Count Subarrays of Length Three With a Condition

Hint 1: Sliding Window Use a sliding window approach to examine each consecutive three-element subarray. You'll need to iterate through the array and check each possible window of size 3.

Hint 2: Mathematical Condition For a window [a, b, c], you need to verify if a + c == b / 2. Be careful with integer division - in Python 3, use // for integer division or ensure proper comparison handling with floating-point division.

Hint 3: Edge Cases Consider arrays with length less than 3 (return 0), arrays with zeros, and negative numbers. The condition should work with standard arithmetic operations.

Full Solution ` def countValidWindows(nums): # If array has fewer than 3 elements, no valid windows exist if len(nums) < 3: return 0

count = 0

# Iterate through all possible windows of size 3
for i in range(len(nums) - 2):
    # Extract the three elements
    first = nums[i]
    middle = nums[i + 1]
    third = nums[i + 2]
    
    # Check if the condition is satisfied: first + third = middle / 2
    # We need to check if middle is divisible by 2 to avoid float comparison issues
    # But actually we can just use direct comparison with division
    if first + third == middle / 2:
        count += 1

return count

`

Explanation:

The solution uses a straightforward sliding window approach:

1. Edge Case Handling: First, we check if the array has fewer than 3 elements. If so, there are no valid windows, so we return 0.
2. Window Iteration: We iterate through the array using index i from 0 to len(nums) - 3 (inclusive). This ensures we can always access elements at positions i, i+1, and i+2.
3. Condition Check: For each window, we extract the three elements and check if first + third == middle / 2. This is the exact mathematical condition specified in the problem.
4. Counting: Each time we find a window that satisfies the condition, we increment our counter.

Time Complexity: O(n) where n is the length of the input array. We make a single pass through the array, examining each window exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for variables (count, first, middle, third).

The algorithm is efficient and handles all edge cases including negative numbers, zeros, and arrays of various sizes.

Condition Check: For each window, we extract the three elements and check if first + third == middle / 2. This is the exact mathematical condition specified in the problem.

Counting: Each time we find a window that satisfies the condition, we increment our counter.

Time Complexity: O(n) where n is the length of the input array. We make a single pass through the array, examining each window exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for variables (count, first, middle, third).

The algorithm is efficient and handles all edge cases including negative numbers, zeros, and arrays of various sizes.