<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:
        nums = list(set(nums))
        num, i = 1, 0
        nums.sort()
        status = 0
        if 1 not in nums:
            return 1
        while nums[i] < 1:
            i += 1
        nums = nums[i:]
        if len(nums) == nums[len(nums)-1]:
            return nums[len(nums)-1] + 1
        min = nums[len(nums)-1]
        left, right = 0, len(nums)-1
        while left < right:
            mid = (left+right)//2
            if len(nums[left:mid+1]) != (nums[mid]-nums[left] + 1):
                if nums[mid] - nums[mid-1] != 1 and min > nums[mid-1]+1:
                    min = nums[mid-1]+1
                right = mid-1
                continue
            if nums[mid+1] - nums[mid] !=  1 and min > nums[mid]+1:
                min = nums[mid]+1
            left = mid+1
        return min
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