<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:
        """
        Returns:
            index i where all elements in nums[:i] <= elements in nums[i:]
            There is at least one valid answer for the given input
        
        Thinking:
            Brute force => test each i O(n) * O(n * n) => O(n**3)
        
            Better sol =>
                Try to solve in O(n) => 
                    1. construct min_after[i] => min val in nums[i:] (build backwards)
                    2. forward update i
        """
        min_after = [math.inf] * len(nums)
        for i in range(len(nums) - 2, -1, -1):
            min_after[i] = min(nums[i + 1], min_after[i + 1])
        min_after[len(nums) - 1] = -math.inf

        max_before = -math.inf
        result = 0
        for i in range(len(nums) - 1):
            max_before = max(max_before, nums[i])
            if max_before <= min_after[i]:
                result = i + 1
                break
        
        return result
        
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