<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, nums: List[int], threshold: int) -> int:
        # Use a monotonic stack
        # For each # in nums, we want to track the index of the first 
        # right and left number less than it. This way we are guaranteed
        # a subarray where that # is the smallest. Finally, for each 
        # subarray, check if # < threshold / k

        # Initialize stacks
        stack = [] # store indicies, monotonicly increasing
        left = []
        right = []
        for _ in range(len(nums)):
            left.append(-1)
            right.append(len(nums))
        
        # for left
        for i in range(len(nums)):
            while stack and nums[i] <= nums[stack[-1]]:
                stack.pop()
            if stack and nums[i] > nums[stack[-1]]:
                left[i] = stack[-1]
            stack.append(i)

        stack.clear()

        # for right stack
        for i in range(len(nums)-1, -1, -1):
            while stack and nums[i] <= nums[stack[-1]]:
                stack.pop()
            if stack and nums[i] > nums[stack[-1]]:
                right[i] = stack[-1]
            stack.append(i)

        # Check condition for all subarrays
        # For each subarray, we are guaranteed that nums[i] is the smallest
        # element
        # x > threshold / k <==> x * k > threshold
        for i in range(len(nums)):
            if nums[i] * (right[i] - left[i] - 1) > threshold:
                return right[i] - left[i] - 1

        return -1


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