<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code>, and a <strong>0-indexed</strong> integer array <code>pattern</code> of size <code>m</code> consisting of integers <code>-1</code>, <code>0</code>, and <code>1</code>.</p>

<p>A <span data-keyword="subarray">subarray</span> <code>nums[i..j]</code> of size <code>m + 1</code> is said to match the <code>pattern</code> if the following conditions hold for each element <code>pattern[k]</code>:</p>

<ul>
	<li><code>nums[i + k + 1] &gt; nums[i + k]</code> if <code>pattern[k] == 1</code>.</li>
	<li><code>nums[i + k + 1] == nums[i + k]</code> if <code>pattern[k] == 0</code>.</li>
	<li><code>nums[i + k + 1] &lt; nums[i + k]</code> if <code>pattern[k] == -1</code>.</li>
</ul>

<p>Return <em>the<strong> count</strong> of subarrays in</em> <code>nums</code> <em>that match the</em> <code>pattern</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6], pattern = [1,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The pattern [1,1] indicates that we are looking for strictly increasing subarrays of size 3. In the array nums, the subarrays [1,2,3], [2,3,4], [3,4,5], and [4,5,6] match this pattern.
Hence, there are 4 subarrays in nums that match the pattern.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,4,1,3,5,5,3], pattern = [1,0,-1]
<strong>Output:</strong> 2
<strong>Explanation: </strong>Here, the pattern [1,0,-1] indicates that we are looking for a sequence where the first number is smaller than the second, the second is equal to the third, and the third is greater than the fourth. In the array nums, the subarrays [1,4,4,1], and [3,5,5,3] match this pattern.
Hence, there are 2 subarrays in nums that match the pattern.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n == nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= m == pattern.length &lt; n</code></li>
	<li><code>-1 &lt;= pattern[i] &lt;= 1</code></li>
</ul>

class Solution:
    def countMatchingSubarrays(self, nums: List[int], pattern: List[int]) -> int:
        n, p = len(nums), len(pattern)
        count = 0

        # helper function to check if the subarray matches the pattern
        def isValidSubArr(arr):
            # check the pattern
            for j in range(1, len(arr)):
                # check if the curr element is greater than
                if pattern[j-1] == 1:
                    if arr[j-1] >= arr[j]:
                        return False
                elif pattern[j-1] == 0:
                    if arr[j-1] != arr[j]:
                        return False
                elif pattern[j-1] == -1:
                    if arr[j-1] <= arr[j]:
                        return False
            return True
        # go through every int in the array up until the last index that has a valid p length subarray
        for i in range(n - p):
            arr = nums[i : i + p + 1]
            isValid = isValidSubArr(arr)
            if isValid: 
                count += 1
        return count
#pay attention


 



        

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