<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 {
public:
    int firstMissingPositive(vector<int>& nums) {

        // Remove all non positive numbers
        nums = removeNeg(nums);

        // Now we have array of size n so for each number in the range n, Mark the number in its index as -1, So when iterating if the index val is -1
        // that means indx+1 is in the array
        // Now get the min indx which is not -1 and return the index+1

        for(auto x: nums){
            int k = abs(x)-1;
            if(k<nums.size() && nums[k]>0){
                nums[k] *= -1;
            }
        }

        for(int i=0; i<nums.size(); i++){
            if(nums[i]>0){
                return i+1;
            }
        }
        return nums.size()+1;
    }

    vector<int> removeNeg(vector<int> vec){
        vector<int> ans;

        for(auto x: vec){
            if(x>0){
                ans.push_back(x);
            }
        }
        return ans;
    }
};
51883