<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:
    std::vector<int> findErrorNums(std::vector<int>& nums) {
        int n = nums.size();
        long long sum = 0, sumSquares = 0;
        long long sumExpected = (long long)n * (n + 1) / 2;
        long long sumSquaresExpected = (long long)n * (n + 1) * (2 * n + 1) / 6;
        
        for (int num : nums) {
            sum += num;
            sumSquares += (long long)num * num;
        }
        
        long long diff = sum - sumExpected;
        long long diffSquares = sumSquares - sumSquaresExpected;
        
        long long missingPlusDuplicated = diffSquares / diff;
        
        int duplicated = (int)((missingPlusDuplicated - diff) / 2);
        int missing = (int)(duplicated + diff);
        
        return {missing, duplicated};
    }
};

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