Subarray With Elements Greater Than Varying Threshold
<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>
Stack; Union Find; Array; Monotonic Stack
using pii = pair<int, int>;

class Solution {
public:
    vector<int> p;
    vector<int> size;

    int validSubarraySize(vector<int>& nums, int threshold) {
        int n = nums.size();
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        size.assign(n, 1);
        vector<pii> arr(n);
        for (int i = 0; i < n; ++i) arr[i] = {nums[i], i};
        sort(arr.begin(), arr.end());
        vector<bool> vis(n);
        for (int j = n - 1; ~j; --j) {
            int v = arr[j].first, i = arr[j].second;
            if (i && vis[i - 1]) merge(i, i - 1);
            if (j < n - 1 && vis[i + 1]) merge(i, i + 1);
            if (v > threshold / size[find(i)]) return size[find(i)];
            vis[i] = true;
        }
        return -1;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }

    void merge(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) return;
        p[pa] = pb;
        size[pb] += size[pa];
    }
};