{ "data": { "question": { "questionId": "2325", "questionFrontendId": "2222", "categoryTitle": "Algorithms", "boundTopicId": 1385553, "title": "Number of Ways to Select Buildings", "titleSlug": "number-of-ways-to-select-buildings", "content": "

You are given a 0-indexed binary string s which represents the types of buildings along a street where:

\n\n\n\n

As a city official, you would like to select 3 buildings for random inspection. However, to ensure variety, no two consecutive buildings out of the selected buildings can be of the same type.

\n\n\n\n

Return the number of valid ways to select 3 buildings.

\n\n

 

\n

Example 1:

\n\n
\nInput: s = "001101"\nOutput: 6\nExplanation: \nThe following sets of indices selected are valid:\n- [0,2,4] from "001101" forms "010"\n- [0,3,4] from "001101" forms "010"\n- [1,2,4] from "001101" forms "010"\n- [1,3,4] from "001101" forms "010"\n- [2,4,5] from "001101" forms "101"\n- [3,4,5] from "001101" forms "101"\nNo other selection is valid. Thus, there are 6 total ways.\n
\n\n

Example 2:

\n\n
\nInput: s = "11100"\nOutput: 0\nExplanation: It can be shown that there are no valid selections.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "选择建筑的方案数", "translatedContent": "

给你一个下标从 0 开始的二进制字符串 s ，它表示一条街沿途的建筑类型，其中：

\n\n\n\n

作为市政厅的官员，你需要随机 选择 3 栋建筑。然而，为了确保多样性，选出来的 3 栋建筑 相邻 的两栋不能是同一类型。

\n\n\n\n

请你返回可以选择 3 栋建筑的 有效方案数 。

\n\n

 

\n\n

示例 1：

\n\n
输入：s = \"001101\"\n输出：6\n解释：\n以下下标集合是合法的：\n- [0,2,4] ，从 \"001101\" 得到 \"010\"\n- [0,3,4] ，从 \"001101\" 得到 \"010\"\n- [1,2,4] ，从 \"001101\" 得到 \"010\"\n- [1,3,4] ，从 \"001101\" 得到 \"010\"\n- [2,4,5] ，从 \"001101\" 得到 \"101\"\n- [3,4,5] ，从 \"001101\" 得到 \"101\"\n没有别的合法选择，所以总共有 6 种方法。\n
\n\n

示例 2：

\n\n
输入：s = \"11100\"\n输出：0\n解释：没有任何符合题意的选择。\n
\n\n

 

\n\n

提示：

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Think about how we can construct these 2 patterns from smaller patterns.", "Count the number of subsequences of the form ‘01’ or ‘10’ first. Let n01[i] be the number of ‘01’ subsequences that exist in the prefix of s up to the ith building. How can you compute n01[i]?", "Let n0[i] and n1[i] be the number of ‘0’s and ‘1’s that exists in the prefix of s up to i respectively. Then n01[i] = n01[i – 1] if s[i] == ‘0’, otherwise n01[i] = n01[i – 1] + n0[i – 1].", "The same logic applies to building the n10 array and subsequently the n101 and n010 arrays for the number of ‘101’ and ‘010‘ subsequences." ], "solution": null, "status": null, "sampleTestCase": "\"001101\"", "metaData": "{\n \"name\": \"numberOfWays\",\n \"params\": [\n {\n \"name\": \"s\",\n \"type\": \"string\"\n }\n ],\n \"return\": {\n \"type\": \"long\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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