<p>Given an unsorted integer array <code>nums</code>. Return the <em>smallest positive integer</em> that is <em>not present</em> in <code>nums</code>.</p>

<p>You must implement an algorithm that runs in <code>O(n)</code> time and uses <code>O(1)</code> auxiliary space.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The numbers in the range [1,2] are all in the array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,4,-1,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 1 is in the array but 2 is missing.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,8,9,11,12]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The smallest positive integer 1 is missing.
</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>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution:
    def firstMissingPositive(self, nums: List[int]) -> int:
        def partition(nums):
            i = -1
            for j in range(len(nums)):
                if nums[j] > 0 and nums[j] <= len(nums):
                    i += 1
                    nums[i], nums[j] = nums[j], nums[j]
            return i

        def left(i):
            return i * 2 + 1

        def right(i):
            return i * 2 + 2

        def min_heapify(A, i, heap_size):
            l = left(i)
            r = right(i)
            if l <= heap_size and A[l] < A[i]:
                smallest = l
            else:
                smallest = i
            if r <= heap_size and A[r] < A[smallest]:
                smallest = r
            if smallest != i:
                A[i], A[smallest] = A[smallest], A[i]
                min_heapify(A, smallest, heap_size)
            
        def build_min_heap(A, heap_size):
            for i in range((heap_size + 1)//2, -1, -1):
                min_heapify(A, i, heap_size)


        last_positive_idx = partition(nums)
        if last_positive_idx == -1:
            return 1
        
        build_min_heap(nums, last_positive_idx)
        def extract_min(A, heap_size):
            min_n = A[0]
            A[0] = A[heap_size]
            heap_size -= 1
            min_heapify(A, 0, heap_size)
            return min_n, heap_size

        if nums[0] != 1:
            return 1
        heap_size = last_positive_idx
        n = 1
        for _ in range(last_positive_idx+1):
            min_n, heap_size = extract_min(nums, heap_size)
            if min_n > n:
                return n
            n = min_n + 1
        return n
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