<p>You have a set of integers <code>s</code>, which originally contains all the numbers from <code>1</code> to <code>n</code>. Unfortunately, due to some error, one of the numbers in <code>s</code> got duplicated to another number in the set, which results in <strong>repetition of one</strong> number and <strong>loss of another</strong> number.</p>

<p>You are given an integer array <code>nums</code> representing the data status of this set after the error.</p>

<p>Find the number that occurs twice and the number that is missing and return <em>them in the form of an array</em>.</p>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums)
    {
           vector<int> result(2,0);

           sort(nums.begin(), nums.end()); // sort

           int fixed_sum = 0;
           for (int i(1); i<=nums.size();++i)
           {
                fixed_sum += i;
           }

           int sum = 0;
           for (int el(0); el < nums.size(); ++el) // search duplicate
           {
              if (el > 0 && nums[el] == nums[el - 1])
              {
                  result[0] = nums[el];
              }
              else
              {
                    sum+= nums[el];
              }

           }
           result[1] = fixed_sum - sum;
           return result;
    }
};
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