2028

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 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.

Example 4:

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

 

Constraints:

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

Similar Questions:

• Number of Dice Rolls With Target Sum (Medium)
• Dice Roll Simulation (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/find-missing-observations/
// Author: github.com/lzl124631x
// 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;
    }
};