<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:
        '''
        作用：
        触发栈清空：0 是一个很小的值，确保在遍历到数组末尾时，会将栈中的所有元素弹出。
        否则，如果栈中还有元素没有弹出，那么对于这些剩余元素，
        栈中的值可能没有机会被处理，导致算法无法遍历所有可能的子数组。

        结束逻辑的简化：通过添加一个 0，我们可以避免在遍历 nums 的过程中再额外处理栈清空的逻辑。
        添加这个最小值可以让算法自动处理栈中剩余的元素，使代码更加简洁。
        '''                  
        nums.append(0)
        stack = deque()

        for idx in range(len(nums)):

            while stack and nums[idx] <= nums[stack[-1]]:           
                n = nums[stack.pop()]  
                print("n:") 
                print(n)
                k = idx if not stack else idx - stack[-1] -1    
                print("k:")
                print(k)    
                if n > threshold //k: return k                     
                                        
            stack.append(idx)

        return -1
32299