<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:
    vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
        int m = rolls.size(), sum = 0;
        for(int roll: rolls)
            sum += roll;
        int rem = mean*(m+n) - sum;
        if(rem > n*6 || rem < n)
            return {};
        int m2 = rem/n;
        if(n*m2 == rem)
            return vector<int>(n, m2);
        else
        {
            int remrem = rem-m2*n;
            vector<int> ans(n-remrem, m2);
            vector<int> ans2(remrem, m2+1);
            ans.insert(ans.end(), ans2.begin(), ans2.end());
            return ans;
        }
    }
};
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