Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 5

Description

Find Missing Observations

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

Solution

class Solution {
public:
  vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
    int sum = accumulate(rolls.begin(), rolls.end(), 0);
    int m = rolls.size();
    if(mean * (m + n) < sum + n || mean * (m + n) > sum + n * 6) return {};
    sum = mean * (m + n) - sum;
    vector<int> answer;
    while(n) {
      int current;
      if(n & 1) {
        current = ceil(sum * 1.0 / n);
      } else {
        current = sum / n;
      }
      n -= 1;
      sum -= current;
      answer.push_back(current);
    }
    return answer;
  }
};