{ "data": { "question": { "questionId": "2155", "questionFrontendId": "2028", "categoryTitle": "Algorithms", "boundTopicId": 1026586, "title": "Find Missing Observations", "titleSlug": "find-missing-observations", "content": "
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
.
\n
Example 1:
\n\n\nInput: rolls = [3,2,4,3], mean = 4, n = 2\nOutput: [6,6]\nExplanation: The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.\n\n\n
Example 2:
\n\n\nInput: rolls = [1,5,6], mean = 3, n = 4\nOutput: [2,3,2,2]\nExplanation: The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.\n\n\n
Example 3:
\n\n\nInput: rolls = [1,2,3,4], mean = 6, n = 4\nOutput: []\nExplanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.\n\n\n
\n
Constraints:
\n\nm == rolls.length
1 <= n, m <= 105
1 <= rolls[i], mean <= 6
现有一份 n + m
次投掷单个 六面 骰子的观测数据,骰子的每个面从 1
到 6
编号。观测数据中缺失了 n
份,你手上只拿到剩余 m
次投掷的数据。幸好你有之前计算过的这 n + m
次投掷数据的 平均值 。
给你一个长度为 m
的整数数组 rolls
,其中 rolls[i]
是第 i
次观测的值。同时给你两个整数 mean
和 n
。
返回一个长度为 n
的数组,包含所有缺失的观测数据,且满足这 n + m
次投掷的 平均值 是 mean
。如果存在多组符合要求的答案,只需要返回其中任意一组即可。如果不存在答案,返回一个空数组。
k
个数字的 平均值 为这些数字求和后再除以 k
。
注意 mean
是一个整数,所以 n + m
次投掷的总和需要被 n + m
整除。
\n\n
示例 1:
\n\n\n输入:rolls = [3,2,4,3], mean = 4, n = 2\n输出:[6,6]\n解释:所有 n + m 次投掷的平均值是 (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4 。\n\n\n
示例 2:
\n\n\n输入:rolls = [1,5,6], mean = 3, n = 4\n输出:[2,3,2,2]\n解释:所有 n + m 次投掷的平均值是 (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3 。\n\n\n
示例 3:
\n\n\n输入:rolls = [1,2,3,4], mean = 6, n = 4\n输出:[]\n解释:无论丢失的 4 次数据是什么,平均值都不可能是 6 。\n\n\n
示例 4:
\n\n\n输入:rolls = [1], mean = 3, n = 1\n输出:[5]\n解释:所有 n + m 次投掷的平均值是 (1 + 5) / 2 = 3 。\n\n\n
\n\n
提示:
\n\nm == rolls.length
1 <= n, m <= 105
1 <= rolls[i], mean <= 6
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