You have observations of n + m6-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 lengthncontaining the missing observations such that the average value of then + mrolls is exactlymean. 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.
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.
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.
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.
m == rolls.length1 <= n, m <= 1051 <= rolls[i], mean <= 6
implSolution{pubfnmissing_rolls(rolls:Vec<i32>,mean:i32,n:i32) -> Vec<i32>{letmut n_sum = mean *(rolls.len()asi32 + n) - rolls.into_iter().sum::<i32>();letmut ret = vec![];if n_sum < n || n_sum > 6* n {return ret;}for i in(0..n).rev(){let x = (n_sum - 6* i).max(1); n_sum -= x; ret.push(x);} ret }}