LeetCode-in-Java
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.
Constraints:
m == rolls.length1 <= n, m <= 1051 <= rolls[i], mean <= 6
Solution
public class Solution {
public int[] missingRolls(int[] rolls, int mean, int n) {
int m = rolls.length;
int msum = 0;
int[] res = new int[n];
for (int roll : rolls) {
msum += roll;
}
int totalmn = mean * (m + n) - msum;
if (totalmn < n || totalmn > n * 6) {
return new int[0];
}
int j = 0;
while (totalmn > 0) {
int dice = Math.min(6, totalmn - n + 1);
res[j] = dice;
totalmn = totalmn - dice;
n--;
j++;
}
return res;
}
}