<p>Given an integer array <code>nums</code>, partition it into two (contiguous) subarrays <code>left</code> and <code>right</code> so that:</p>

<ul>
	<li>Every element in <code>left</code> is less than or equal to every element in <code>right</code>.</li>
	<li><code>left</code> and <code>right</code> are non-empty.</li>
	<li><code>left</code> has the smallest possible size.</li>
</ul>

<p>Return <em>the length of </em><code>left</code><em> after such a partitioning</em>.</p>

<p>Test cases are generated such that partitioning exists.</p>

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

<pre>
<strong>Input:</strong> nums = [5,0,3,8,6]
<strong>Output:</strong> 3
<strong>Explanation:</strong> left = [5,0,3], right = [8,6]
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,0,6,12]
<strong>Output:</strong> 4
<strong>Explanation:</strong> left = [1,1,1,0], right = [6,12]
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li>There is at least one valid answer for the given input.</li>
</ul>

class Solution:
    def partitionDisjoint(self, nums: List[int]) -> int:
        supp = [[0, 0] for x in range(len(nums)-1)]
        cur = 0
        for i, v in enumerate(nums[:-1]):
            cur = max(cur, v)
            supp[i][0] = cur
        for i, v in enumerate(nums[:0:-1]):
            cur = min(cur, v)
            supp[-i-1][1] = cur
        for i, (a, b) in enumerate(supp):
            if a <= b:
                return i + 1
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