# 2028. Find Missing Observations

## Description

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 ith observation. You are also given the two integers 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 = 2
Output: [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 = 4
Output: [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 = 4
Output: []
Explanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.


Constraints:

• m == rolls.length
• 1 <= n, m <= 105
• 1 <= rolls[i], mean <= 6

## Solutions

Solution 1: Construction

According to the problem description, the sum of all numbers is $(n + m) \times mean$, and the sum of known numbers is sum(rolls). Therefore, the sum of the missing numbers is $s = (n + m) \times mean - sum(rolls)$.

If $s > n \times 6$ or $s < n$, it means there is no answer that satisfies the conditions, so return an empty array.

Otherwise, we can evenly distribute $s$ to $n$ numbers, that is, the value of each number is $s / n$, and the value of $s \bmod n$ numbers is increased by $1$.

The time complexity is $O(n + m)$, and the space complexity is $O(1)$. Here, $n$ and $m$ are the number of missing numbers and known numbers, respectively.

• class Solution {
public int[] missingRolls(int[] rolls, int mean, int n) {
int m = rolls.length;
int s = (n + m) * mean;
for (int v : rolls) {
s -= v;
}
if (s > n * 6 || s < n) {
return new int[0];
}
int[] ans = new int[n];
Arrays.fill(ans, s / n);
for (int i = 0; i < s % n; ++i) {
++ans[i];
}
return ans;
}
}

• class Solution {
public:
vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
int m = rolls.size();
int s = (n + m) * mean;
for (int& v : rolls) s -= v;
if (s > n * 6 || s < n) return {};
vector<int> ans(n, s / n);
for (int i = 0; i < s % n; ++i) ++ans[i];
return ans;
}
};

• class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
m = len(rolls)
s = (n + m) * mean - sum(rolls)
if s > n * 6 or s < n:
return []
ans = [s // n] * n
for i in range(s % n):
ans[i] += 1
return ans


• func missingRolls(rolls []int, mean int, n int) []int {
m := len(rolls)
s := (n + m) * mean
for _, v := range rolls {
s -= v
}
if s > n*6 || s < n {
return []int{}
}
ans := make([]int, n)
for i, j := 0, 0; i < n; i, j = i+1, j+1 {
ans[i] = s / n
if j < s%n {
ans[i]++
}
}
return ans
}

• function missingRolls(rolls: number[], mean: number, n: number): number[] {
const len = rolls.length + n;
const sum = rolls.reduce((p, v) => p + v);
const max = n * 6;
const min = n;
if ((sum + max) / len < mean || (sum + min) / len > mean) {
return [];
}

const res = new Array(n);
for (let i = min; i <= max; i++) {
if ((sum + i) / len === mean) {
const num = Math.floor(i / n);
res.fill(num);
let count = i - n * num;
let j = 0;
while (count != 0) {
if (res[j] === 6) {
j++;
} else {
res[j]++;
count--;
}
}
break;
}
}
return res;
}


• impl Solution {
pub fn missing_rolls(rolls: Vec<i32>, mean: i32, n: i32) -> Vec<i32> {
let n = n as usize;
let mean = mean as usize;
let len = rolls.len() + n;
let sum: i32 = rolls.iter().sum();
let sum = sum as usize;
let max = n * 6;
let min = n;
if sum + max < mean * len || sum + min > mean * len {
return vec![];
}

let mut res = vec![0; n];
for i in min..=max {
if (sum + i) / len == mean {
let num = i / n;
res.fill(num as i32);
let mut count = i - n * num;
let mut j = 0;
while count != 0 {
if res[j] == 6 {
j += 1;
} else {
res[j] += 1;
count -= 1;
}
}
break;
}
}
res
}
}