<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:
        answer = 0 
        n = len(nums)
        left_max = n*[0]
        right_min = n*[0]
        left_max[0] = nums[0]
        iter_ = enumerate(nums)
        next(iter_)
        for idx,i in iter_:
            last_ele = left_max[idx-1]
            left_max[idx] = i if i > last_ele else last_ele
        right_min[-1] =nums[-1]
        reverse_iter = reversed(list(enumerate(nums)))
        next(reverse_iter)
        for idx,i in reverse_iter:
            last_ele = right_min[idx+1]
            right_min[idx] = i if i < last_ele else last_ele      
        iter_ = enumerate(nums)
        answer = 0
        for idx,i in iter_:
            if idx < n-1:
                if left_max[idx] <= right_min[idx+1]:
                    return idx+1
            else:
                return idx 



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