<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>


#include <span>
#include <array>
#include <vector>
using namespace std;

/*
 The code will run faster with ios::sync_with_stdio(0).
 However, this should not be used in production code and interactive problems.
 In this particular problem, it is ok to apply ios::sync_with_stdio(0).

 Many of the top-ranked C++ solutions for time on leetcode apply this trick,
 so, for a fairer assessment of the time percentile of my code
 you could outcomment the lambda expression below for a faster run.
*/

/*
 const static auto speedup = [] {
	ios::sync_with_stdio(0);
	return nullptr;
 }();
*/

class Solution {

    static constexpr array<int, 2> RANGE_OF_VALUES{ { 1, 6 } };
    inline static const vector<int> NO_VALID_ANSWER;

public:
    vector<int> missingRolls(const vector<int>& rolls, int mean, int numberOfMissingObservations) const {
        int sumOfMissingObservations = calculateSumOfMissingObservations(rolls, mean, numberOfMissingObservations);

        if (sumOfMissingObservations < numberOfMissingObservations * RANGE_OF_VALUES[0]
            || sumOfMissingObservations > numberOfMissingObservations * RANGE_OF_VALUES[1]) {
            return NO_VALID_ANSWER;
        }

        return extractMissingValues(sumOfMissingObservations, numberOfMissingObservations);
    }

private:
    int calculateSumOfMissingObservations(span <const int> rolls, int mean, int numberOfMissingObservations) const {
        int sumObservedValues = 0;
        for (int value : rolls) {
            sumObservedValues += value;
        }
        return mean * (rolls.size() + numberOfMissingObservations) - sumObservedValues;
    }

    vector<int> extractMissingValues(int sumOfMissingObservations, int numberOfMissingObservations) const {
        vector<int> missingValues(numberOfMissingObservations);
        int roundedDownMeanOfMissingObsarvations = sumOfMissingObservations / numberOfMissingObservations;
        int remainder = sumOfMissingObservations % numberOfMissingObservations;

        for (auto& missingValue : missingValues) {
            missingValue = roundedDownMeanOfMissingObsarvations;
            if (remainder > 0) {
                ++missingValue;
                --remainder;
            }
        }
        return missingValues;
    }
};
22060