<p>You have observations of <code>n + m</code> <strong>6-sided</strong> dice rolls with each face numbered from <code>1</code> to <code>6</code>. <code>n</code> of the observations went missing, and you only have the observations of <code>m</code> rolls. Fortunately, you have also calculated the <strong>average value</strong> of the <code>n + m</code> rolls.</p>

<p>You are given an integer array <code>rolls</code> of length <code>m</code> where <code>rolls[i]</code> is the value of the <code>i<sup>th</sup></code> observation. You are also given the two integers <code>mean</code> and <code>n</code>.</p>

<p>Return <em>an array of length </em><code>n</code><em> containing the missing observations such that the <strong>average value </strong>of the </em><code>n + m</code><em> rolls is <strong>exactly</strong> </em><code>mean</code>. If there are multiple valid answers, return <em>any of them</em>. If no such array exists, return <em>an empty array</em>.</p>

<p>The <strong>average value</strong> of a set of <code>k</code> numbers is the sum of the numbers divided by <code>k</code>.</p>

<p>Note that <code>mean</code> is an integer, so the sum of the <code>n + m</code> rolls should be divisible by <code>n + m</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> rolls = [3,2,4,3], mean = 4, n = 2
<strong>Output:</strong> [6,6]
<strong>Explanation:</strong> The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> rolls = [1,5,6], mean = 3, n = 4
<strong>Output:</strong> [2,3,2,2]
<strong>Explanation:</strong> The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> rolls = [1,2,3,4], mean = 6, n = 4
<strong>Output:</strong> []
<strong>Explanation:</strong> It is impossible for the mean to be 6 no matter what the 4 missing rolls are.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == rolls.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= rolls[i], mean &lt;= 6</code></li>
</ul>

class Solution {
public:
    Solution() {
        std::ios_base::sync_with_stdio(false);
        std::cin.tie(NULL);
        std::cout.tie(NULL);
    }

    vector<int> missingRolls(vector<int>& rolls, int mean, int n) 
    {
        const int MAX_missing_sum = n * 6;
        const int MIN_missing_sum = n;
        int n0 = rolls.size();
        int sum = accumulate(rolls.begin(), rolls.end(), 0);
        int og_n = n0 + n;
        int og_sum = mean * (og_n);

        int sum_of_missing = og_sum - sum;

        if ((sum_of_missing > MAX_missing_sum) || (sum_of_missing < MIN_missing_sum))
            return {};

        int div = sum_of_missing / n;
        int rem = sum_of_missing % n;

        vector<int> missing(n, div);

        if (rem == 0) {
            return missing;
        } 
        else 
        {
            for (auto& i : missing) 
            {
                if (i + rem > 6) 
                {
                    rem = rem - (6 - i);
                    i = 6;
                } 
                else 
                {
                    i = i + rem;
                    return missing;
                }
            }
        }

        return {-1};
    }
};
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