Formatted question description: https://leetcode.ca/all/2028.html

# 2028. Find Missing Observations (Medium)

You have observations of `n + m`

**6-sided** dice rolls with each face numbered from `1`

to `6`

. `n`

of the observations went missing, and you only have the observations of `m`

rolls. Fortunately, you have also calculated the **average value** of the `n + m`

rolls.

You are given an integer array `rolls`

of length `m`

where `rolls[i]`

is the value of the `i`

observation. You are also given the two integers ^{th}`mean`

and `n`

.

Return *an array of length *`n`

* containing the missing observations such that the average value of the *

`n + m`

*rolls is*

**exactly**`mean`

. If there are multiple valid answers, return *any of them*. If no such array exists, return

*an empty array*.

The **average value** of a set of `k`

numbers is the sum of the numbers divided by `k`

.

Note that `mean`

is an integer, so the sum of the `n + m`

rolls should be divisible by `n + m`

.

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

**Example 4:**

Input:rolls = [1], mean = 3, n = 1Output:[5]Explanation:The mean of all n + m rolls is (1 + 5) / 2 = 3.

**Constraints:**

`m == rolls.length`

`1 <= n, m <= 10`

^{5}`1 <= rolls[i], mean <= 6`

**Similar Questions**:

## Solution 1.

```
// OJ: https://leetcode.com/problems/find-missing-observations/
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
vector<int> missingRolls(vector<int>& A, int mean, int n) {
int m = A.size(), sum = accumulate(begin(A), end(A), 0), goal = (n + m) * mean - sum;
if (goal < n || goal > 6 * n) return {};
vector<int> ans;
int d = goal / n, r = goal % n;
for (int i = 0; i < n; ++i) {
ans.push_back(d + (r > 0));
--r;
}
return ans;
}
};
```