<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) {
        sort(nums.begin(),nums.end());
        vector <int> ans;
        vector <int> nums2(nums.size()+1,0);
        for(int i=0;i<nums.size();i++){
            nums2[nums[i]]++;
        }
        for(int i=1;i<nums2.size();i++){
            if(nums2[i]==0){
                ans.push_back(i);
            }
        }
        return ans;
    }
};
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