<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<int> num;
        int l=nums.size(), j=0;
        //cout << "size: " << l << "\n";
        cout << "length of nums: " << l << "\n";
        for (int i=0; i<l; i++) { if (nums[i] > 0) { num.insert ( num.end() , nums[i] ); }}
        vector<int> myarray;
        l=num.size(); cout << "length of num: " << l << "\n";
        for (int i=0; i<=l; i++) { myarray.insert ( myarray.end() , i+1); } // populate myarray
        for (int i=0; i<l; i++) { if (num[i]<=l) {myarray[num[i]-1]=0;}} // manipulate myarray
        for (int i=0; i<=l; i++) { if (myarray[i]>0) {return i+1; }}
    return 0;
    }
};
58973