{ "data": { "question": { "questionId": "2390", "questionFrontendId": "2306", "categoryTitle": "Algorithms", "boundTopicId": 1592196, "title": "Naming a Company", "titleSlug": "naming-a-company", "content": "

You are given an array of strings ideas that represents a list of names to be used in the process of naming a company. The process of naming a company is as follows:

\n\n
    \n\t
  1. Choose 2 distinct names from ideas, call them ideaA and ideaB.
  2. \n\t
  3. Swap the first letters of ideaA and ideaB with each other.
  4. \n\t
  5. If both of the new names are not found in the original ideas, then the name ideaA ideaB (the concatenation of ideaA and ideaB, separated by a space) is a valid company name.
  6. \n\t
  7. Otherwise, it is not a valid name.
  8. \n
\n\n

Return the number of distinct valid names for the company.

\n\n

 

\n

Example 1:

\n\n
\nInput: ideas = ["coffee","donuts","time","toffee"]\nOutput: 6\nExplanation: The following selections are valid:\n- ("coffee", "donuts"): The company name created is "doffee conuts".\n- ("donuts", "coffee"): The company name created is "conuts doffee".\n- ("donuts", "time"): The company name created is "tonuts dime".\n- ("donuts", "toffee"): The company name created is "tonuts doffee".\n- ("time", "donuts"): The company name created is "dime tonuts".\n- ("toffee", "donuts"): The company name created is "doffee tonuts".\nTherefore, there are a total of 6 distinct company names.\n\nThe following are some examples of invalid selections:\n- ("coffee", "time"): The name "toffee" formed after swapping already exists in the original array.\n- ("time", "toffee"): Both names are still the same after swapping and exist in the original array.\n- ("coffee", "toffee"): Both names formed after swapping already exist in the original array.\n
\n\n

Example 2:

\n\n
\nInput: ideas = ["lack","back"]\nOutput: 0\nExplanation: There are no valid selections. Therefore, 0 is returned.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "公司命名", "translatedContent": "

给你一个字符串数组 ideas 表示在公司命名过程中使用的名字列表。公司命名流程如下:

\n\n
    \n\t
  1. ideas 中选择 2 个 不同 名字,称为 ideaAideaB
  2. \n\t
  3. 交换 ideaAideaB 的首字母。
  4. \n\t
  5. 如果得到的两个新名字 不在 ideas 中,那么 ideaA ideaB串联 ideaAideaB ,中间用一个空格分隔)是一个有效的公司名字。
  6. \n\t
  7. 否则,不是一个有效的名字。
  8. \n
\n\n

返回 不同 且有效的公司名字的数目。

\n\n

 

\n\n

示例 1:

\n\n
输入:ideas = [\"coffee\",\"donuts\",\"time\",\"toffee\"]\n输出:6\n解释:下面列出一些有效的选择方案:\n- (\"coffee\", \"donuts\"):对应的公司名字是 \"doffee conuts\" 。\n- (\"donuts\", \"coffee\"):对应的公司名字是 \"conuts doffee\" 。\n- (\"donuts\", \"time\"):对应的公司名字是 \"tonuts dime\" 。\n- (\"donuts\", \"toffee\"):对应的公司名字是 \"tonuts doffee\" 。\n- (\"time\", \"donuts\"):对应的公司名字是 \"dime tonuts\" 。\n- (\"toffee\", \"donuts\"):对应的公司名字是 \"doffee tonuts\" 。\n因此,总共有 6 个不同的公司名字。\n\n下面列出一些无效的选择方案:\n- (\"coffee\", \"time\"):在原数组中存在交换后形成的名字 \"toffee\" 。\n- (\"time\", \"toffee\"):在原数组中存在交换后形成的两个名字。\n- (\"coffee\", \"toffee\"):在原数组中存在交换后形成的两个名字。\n
\n\n

示例 2:

\n\n
输入:ideas = [\"lack\",\"back\"]\n输出:0\n解释:不存在有效的选择方案。因此,返回 0 。\n
\n\n

 

\n\n

提示:

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"CodeSnippetNode" }, { "lang": "Kotlin", "langSlug": "kotlin", "code": "class Solution {\n fun distinctNames(ideas: Array): Long {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Dart", "langSlug": "dart", "code": "class Solution {\n int distinctNames(List ideas) {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Go", "langSlug": "golang", "code": "func distinctNames(ideas []string) int64 {\n \n}", "__typename": "CodeSnippetNode" }, { "lang": "Ruby", "langSlug": "ruby", "code": "# @param {String[]} ideas\n# @return {Integer}\ndef distinct_names(ideas)\n \nend", "__typename": "CodeSnippetNode" }, { "lang": "Scala", "langSlug": "scala", "code": "object Solution {\n def distinctNames(ideas: Array[String]): Long = {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Rust", "langSlug": "rust", "code": "impl Solution {\n pub fn distinct_names(ideas: Vec) -> i64 {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Racket", "langSlug": "racket", "code": 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