<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:
    def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
        # m = len(rolls)
        # s = sum(rolls)
        # # mean = (missing+s)/(m+n)
        # missing = mean*(m+n)-s
        missing = mean*(len(rolls)+n)-sum(rolls)
        # check = (missing+s)/(m+n)
        # device missing into n pecies, each less than 7
        # n_quotient = missing / n
        # n_remainder = missing % n
        # print((m,s,missing,n_quotient,n_remainder))
        # if n_quotient > 6 or missing <= 0 or n_quotient < 1:
        #     return []
        if not n <= missing <= 6 * n:
            return []
        
        q, r = divmod(missing, n)
        if not r:
            return n * [q]
        else:
            return r*[q+1] + (n-r)*[q]
        # i = 0
        # while n_remainder != 0:
        #     if i > n:
        #         return []
        #     result[i] += 1
        #     n_remainder -= 1
        #     i += 1
        # return result
        # result = 6*[0]

        #low eff only happen when larger than 6, so not using while loop
        # if missing > 6:
        #     missing = 
        # quotient = a // b   # gives how many times a divides b
        # remainder = a % b   # gives the remainder of the division
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