<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> ans;
vector<int> frq(20000,0);

for(int i = 0;i<nums.size();i++ ){
    frq[nums[i]]++;
}
//  finding error 

    for(int i = 0;i<frq.size();i++)

        {
            
            if(frq[i] == 2)
                {

                    ans.push_back(i);

                    for(int j = 1;j<frq.size();j++)
                    {
                        if(frq[j] == 0)
                        {
                            ans.push_back(j);
                            return ans;
                        }
                    }
                  
                }
       
        }
        return ans;
    }
        
};
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