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

        for (int i = 0; i < nums.size(); i++) {
            if (abs(nums[i]) > 0) {
                int idx = abs(nums[i]) - 1;
                if (idx < nums.size()) {
                    if (nums[idx] > 0) {
                        nums[idx] = -nums[idx];
                    } else if (nums[idx] == 0) {
                        nums[idx] = -1 * (nums.size()+1);
                    }
                }
            }
        }

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