{ "data": { "question": { "questionId": "3224", "questionFrontendId": "2954", "categoryTitle": "Algorithms", "boundTopicId": 2549327, "title": "Count the Number of Infection Sequences", "titleSlug": "count-the-number-of-infection-sequences", "content": "

You are given an integer n and an array sick sorted in increasing order, representing positions of infected people in a line of n people.

\n\n

At each step, one uninfected person adjacent to an infected person gets infected. This process continues until everyone is infected.

\n\n

An infection sequence is the order in which uninfected people become infected, excluding those initially infected.

\n\n

Return the number of different infection sequences possible, modulo 109+7.

\n\n

 

\n

Example 1:

\n\n
\n

Input: n = 5, sick = [0,4]

\n\n

Output: 4

\n\n

Explanation:

\n\n

There is a total of 6 different sequences overall.

\n\n\n
\n\n

Example 2:

\n\n
\n

Input: n = 4, sick = [1]

\n\n

Output: 3

\n\n

Explanation:

\n\n

There is a total of 6 different sequences overall.

\n\n\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "统计感冒序列的数目", "translatedContent": "

给你一个整数 n 和一个下标从 0 开始的整数数组 sick ，数组按 升序 排序。

\n\n

有 n 位小朋友站成一排，按顺序编号为 0 到 n - 1 。数组 sick 包含一开始得了感冒的小朋友的位置。如果位置为 i 的小朋友得了感冒，他会传染给下标为 i - 1 或者 i + 1 的小朋友，前提 是被传染的小朋友存在且还没有得感冒。每一秒中， 至多一位 还没感冒的小朋友会被传染。

\n\n

经过有限的秒数后，队列中所有小朋友都会感冒。感冒序列 指的是 所有 一开始没有感冒的小朋友最后得感冒的顺序序列。请你返回所有感冒序列的数目。

\n\n

由于答案可能很大，请你将答案对 109 + 7 取余后返回。

\n\n

注意，感冒序列 不 包含一开始就得了感冒的小朋友的下标。

\n\n

 

\n\n

示例 1：

\n\n
\n输入：n = 5, sick = [0,4]\n输出：4\n解释：一开始，下标为 1 ，2 和 3 的小朋友没有感冒。总共有 4 个可能的感冒序列：\n- 一开始，下标为 1 和 3 的小朋友可以被传染，因为他们分别挨着有感冒的小朋友 0 和 4 ，令下标为 1 的小朋友先被传染。\n然后，下标为 2 的小朋友挨着感冒的小朋友 1 ，下标为 3 的小朋友挨着感冒的小朋友 4 ，两位小朋友都可以被传染，令下标为 2 的小朋友被传染。\n最后，下标为 3 的小朋友被传染，因为他挨着感冒的小朋友 2 和 4 ，感冒序列为 [1,2,3] 。\n- 一开始，下标为 1 和 3 的小朋友可以被传染，因为他们分别挨着感冒的小朋友 0 和 4 ，令下标为 1 的小朋友先被传染。\n然后，下标为 2 的小朋友挨着感冒的小朋友 1 ，下标为 3 的小朋友挨着感冒的小朋友 4 ，两位小朋友都可以被传染，令下标为 3 的小朋友被传染。\n最后，下标为 2 的小朋友被传染，因为他挨着感冒的小朋友 1 和 3 ，感冒序列为  [1,3,2] 。\n- 感冒序列 [3,1,2] ，被传染的顺序：[0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] 。\n- 感冒序列 [3,2,1] ，被传染的顺序：[0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] 。\n
\n\n

示例 2：

\n\n
\n输入：n = 4, sick = [1]\n输出：3\n解释：一开始，下标为 0 ，2 和 3 的小朋友没有感冒。总共有 3 个可能的感冒序列：\n- 感冒序列 [0,2,3] ，被传染的顺序：[0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3] 。\n- 感冒序列 [2,0,3] ，被传染的顺序：[0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3] 。\n- 感冒序列 [2,3,0] ，被传染的顺序：[0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3] 。\n
\n\n

 

\n\n

提示：

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Then if i == 0 or j == n - 1, each second there is only one kid that can be infected (which is at the other endpoint).", "If i > 0 and j < n - 1, we have two choices per second since the children at the two endpoints can both be the infect candidates. So there are 2j - i orders to infect all children in the segment.", "Each second we can select a segment and select one endpoint from it.", "The answer is: \r\nS! / (len[1]! * len[2]! * ... * len[m]! * lenstart! * lenend!) * 2k \r\nwhere len[1], len[2], ..., len[m] are the lengths of each segment of non-infected children that have an infected child at both endpoints, lenstart and lenend denote the number of non-infected children with infected child at one endpoint, S is the total length of all segments of non-infected children, and k = (len[1] - 1) + (len[2] - 1) + ... + (len[m] - 1)." ], "solution": null, "status": null, "sampleTestCase": "5\n[0,4]", "metaData": "{\n \"name\": \"numberOfSequence\",\n \"params\": [\n {\n \"name\": \"n\",\n \"type\": \"integer\"\n },\n {\n \"type\": \"integer[]\",\n \"name\": \"sick\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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