<p>You are given an integer array <code>nums</code> and an integer <code>threshold</code>.</p>

<p>Find any subarray of <code>nums</code> of length <code>k</code> such that <strong>every</strong> element in the subarray is <strong>greater</strong> than <code>threshold / k</code>.</p>

<p>Return<em> the <strong>size</strong> of <strong>any</strong> such subarray</em>. If there is no such subarray, return <code>-1</code>.</p>

<p>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,4,3,1], threshold = 6
<strong>Output:</strong> 3
<strong>Explanation:</strong> The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.
</pre>

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

<pre>
<strong>Input:</strong> nums = [6,5,6,5,8], threshold = 7
<strong>Output:</strong> 1
<strong>Explanation:</strong> The subarray [8] has a size of 1, and 8 &gt; 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. 
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], threshold &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def validSubarraySize(self, a: List[int], threshold: int) -> int:
        n = len(a)
        
        # Initialize stacks and arrays to store next and previous smaller elements
        stk = []  # used as a stack
        nextS = [-1] * n  # used for storing next smaller elements
        prevS = [-1] * n  # used for storing previous smaller elements
        
        # Find out the next smaller elements for each index
        for i in range(n):
            while stk and a[i] < a[stk[-1]]:
                nextS[stk.pop()] = i
            stk.append(i)
        
        # Clear the stack for reuse
        stk.clear()
        
        # Find out the previous smaller elements for each index
        for i in range(n - 1, -1, -1):
            while stk and a[i] < a[stk[-1]]:
                prevS[stk.pop()] = i
            stk.append(i)
        
        # Check each element to find the valid subarray size
        for i in range(n):
            left = prevS[i]  # left boundary
            right = nextS[i] if nextS[i] != -1 else n  # right boundary
            
            # Length of subarray formed with a[i] as the minimum element
            length = right - left - 1
            
            # Check if the minimum element of this subarray makes it valid
            if a[i] > (threshold / length):
                return length
        
        return -1
31600