<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:
        n = len(nums)
        for i in range(n):
            if nums[i] < 1 or nums[i] > n:
                nums[i] = n + 1
        print(nums)

        for i in range(n):
            val = abs(nums[i])
            if val > n:
                continue
            val -= 1  #since the list is zero indexed,so every value will be at position val - 1
            
            if nums[val] > 0: 
                # For similar numbers, it will keep on fluctuating between negative and positive 
				# which is not our motive here.
                
                nums[val] = -1 * nums[val]

        print(nums)
        for i in range(n):
            if nums[i] >=0:
                return (i + 1) # bcoz list is zero indexed
        
        '''Step 4 -> We will encounter this if no positives were found. This means that all the 
			         numbers are in the range [1....n]. So the missing positive number will be n+1'''
        
        return (n + 1)        
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