<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) {
        sort (nums.begin(), nums.end());

        vector<bool> B(nums.size(), false);

        int a = 0;
        int b = 0;

        int i = 0;

        for (i = 0; i < nums.size() - 1; ++i)
        {
            B[nums[i] - 1] = true;
            if (nums[i] == nums[i + 1])
            {
                a = nums[i];
            }
        }

        for (i = i; i < nums.size(); ++i)
        {
            B[nums[i] - 1] = true;
        }

        for (i = 0; i < B.size(); ++i)
        {
            if (B[i] == false)
            {
                return {a, i+1};
            }
        }
        return {a,555};
       
    }
};

// https://leetcode.com/problems/set-mismatch/
25400