<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) {
        vector<long long>nums2;
        sort(nums.begin(),nums.end());
        for(int i=0;i<nums.size();i++){
            if(nums[i]>0){
                nums2.push_back(nums[i]);
            }
            // else{
            //     i++;
            // }
        }
        // int j=nums[0];
        int ans=0;
        // if(nums.size()==1&&nums[0]==0)return ans;
        for(int i=0;i<nums2.size();i++){
            // if(nums[i]!=i){
            //     ans= i;
            // }
            cout<<nums2[i]<<" ";
        }
        unique(nums2.begin(),nums2.end());
        for(int i=0;i<nums2.size();i++){
            if((i+1)!=nums2[i]){
            ans=i;
            break;
            }

            else{
            ans=nums[nums.size()-1];
            }
            // else 
            // ans=1;
        }
        return ans+1;

    }
};
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