You are given three integers x, y, and z.
You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concactenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring.
Return the maximum possible length of the new string.
\n\nA substring is a contiguous non-empty sequence of characters within a string.
\n\n\n
Example 1:
\n\n\nInput: x = 2, y = 5, z = 1\nOutput: 12\nExplanation: We can concactenate the strings "BB", "AA", "BB", "AA", "BB", and "AB" in that order. Then, our new string is "BBAABBAABBAB". \nThat string has length 12, and we can show that it is impossible to construct a string of longer length.\n\n\n
Example 2:
\n\n\nInput: x = 3, y = 2, z = 2\nOutput: 14\nExplanation: We can concactenate the strings "AB", "AB", "AA", "BB", "AA", "BB", and "AA" in that order. Then, our new string is "ABABAABBAABBAA". \nThat string has length 14, and we can show that it is impossible to construct a string of longer length.\n\n\n
\n
Constraints:
\n\n1 <= x, y, z <= 50给你三个整数 x ，y 和 z 。
这三个整数表示你有 x 个 \"AA\" 字符串，y 个 \"BB\" 字符串，和 z 个 \"AB\" 字符串。你需要选择这些字符串中的部分字符串（可以全部选择也可以一个都不选择），将它们按顺序连接得到一个新的字符串。新字符串不能包含子字符串 \"AAA\" 或者 \"BBB\" 。
请你返回新字符串的最大可能长度。
\n\n子字符串 是一个字符串中一段连续 非空 的字符序列。
\n\n\n\n
示例 1：
\n\n\n输入：x = 2, y = 5, z = 1\n输出：12\n解释： 我们可以按顺序连接 \"BB\" ，\"AA\" ，\"BB\" ，\"AA\" ，\"BB\" 和 \"AB\" ，得到新字符串 \"BBAABBAABBAB\" 。\n字符串长度为 12 ，无法得到一个更长的符合题目要求的字符串。\n\n\n
示例 2：
\n\n\n输入：x = 3, y = 2, z = 2\n输出：14\n解释：我们可以按顺序连接 \"AB\" ，\"AB\" ，\"AA\" ，\"BB\" ，\"AA\" ，\"BB\" 和 \"AA\" ，得到新字符串 \"ABABAABBAABBAA\" 。\n字符串长度为 14 ，无法得到一个更长的符合题目要求的字符串。\n\n\n
\n\n
提示：
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collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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\\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\" \\u7248\\u672c\\uff1a \\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\" \\u7248\\u672c\\uff1a \\u652f\\u6301 https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982array<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002 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