<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:
    for i in range(len(nums)):
      if nums[i] < 0 or nums[i] > len(nums):
        nums[i] = 0
    # after that, only number between 1~len(nums) remains
    # print(len(nums))
    # for i,n in enumerate(nums):
    #   print(i,'->',n)
    
    mark = - (len(nums) + 1)

    for i in range(len(nums)):
      n = nums[i]
      exist = False

      if n == 0 or n == mark:
        continue

      if n < 0:
        n = -n
        exist = True
      n -= 1
      
      # point to self
      if n == i:
        exist = True

      if n < len(nums):
        if nums[n] > 0:
          nums[n] = -nums[n]
        elif nums[n] == 0:
          nums[n] = mark
      
      if exist:
        nums[i] = mark
    
    # for i,n in enumerate(nums):
    #   print(i,'->',n)

    for i in range(len(nums)):
      if nums[i] >= 0:
        return i + 1
    return len(nums) + 1
    
      
        
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