<p>Given an array <code>nums</code> of <code>n</code> integers where <code>nums[i]</code> is in the range <code>[1, n]</code>, return <em>an array of all the integers in the range</em> <code>[1, n]</code> <em>that do not appear in</em> <code>nums</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [4,3,2,7,8,2,3,1]
<strong>Output:</strong> [5,6]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,1]
<strong>Output:</strong> [2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= n</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do it without extra space and in <code>O(n)</code> runtime? You may assume the returned list does not count as extra space.</p>

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        std::vector<int> values;
        for(int i = 0; i <= nums.size(); ++i){
            values.push_back(i);
        }
        for(const auto val:nums){
            values[val] = 0;
        }
        size_t idx_write{0};
        for(auto idx_read{0}; idx_read <= nums.size(); ++idx_read){
            if(values[idx_read] != 0){
                values[idx_write] = values[idx_read];
                ++idx_write;
            }
        }
        values.resize(idx_write);
        return values;
    }
};
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