You are given two strings of the same length s1 and s2 and a string baseStr.
We say s1[i] and s2[i] are equivalent characters.
s1 = "abc" and s2 = "cde", then we have 'a' == 'c', 'b' == 'd', and 'c' == 'e'.Equivalent characters follow the usual rules of any equivalence relation:
\n\n'a' == 'a'.'a' == 'b' implies 'b' == 'a'.'a' == 'b' and 'b' == 'c' implies 'a' == 'c'.For example, given the equivalency information from s1 = "abc" and s2 = "cde", "acd" and "aab" are equivalent strings of baseStr = "eed", and "aab" is the lexicographically smallest equivalent string of baseStr.
Return the lexicographically smallest equivalent string of baseStr by using the equivalency information from s1 and s2.
\n
Example 1:
\n\n\nInput: s1 = "parker", s2 = "morris", baseStr = "parser"\nOutput: "makkek"\nExplanation: Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i].\nThe characters in each group are equivalent and sorted in lexicographical order.\nSo the answer is "makkek".\n\n\n
Example 2:
\n\n\nInput: s1 = "hello", s2 = "world", baseStr = "hold"\nOutput: "hdld"\nExplanation: Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r].\nSo only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld".\n\n\n
Example 3:
\n\n\nInput: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"\nOutput: "aauaaaaada"\nExplanation: We group the equivalent characters in s1 and s2 as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in baseStr except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".\n\n\n
\n
Constraints:
\n\n1 <= s1.length, s2.length, baseStr <= 1000s1.length == s2.lengths1, s2, and baseStr consist of lowercase English letters.给出长度相同的两个字符串s1 和 s2 ，还有一个字符串 baseStr 。
其中 s1[i] 和 s2[i] 是一组等价字符。
s1 = \"abc\" 且 s2 = \"cde\"，那么就有 'a' == 'c', 'b' == 'd', 'c' == 'e'。等价字符遵循任何等价关系的一般规则：
\n\n'a' == 'a''a' == 'b' 则必定有 'b' == 'a''a' == 'b' 且 'b' == 'c' 就表明 'a' == 'c'例如， s1 = \"abc\" 和 s2 = \"cde\" 的等价信息和之前的例子一样，那么 baseStr = \"eed\" , \"acd\" 或 \"aab\"，这三个字符串都是等价的，而 \"aab\" 是 baseStr 的按字典序最小的等价字符串
利用 s1 和 s2 的等价信息，找出并返回 baseStr 的按字典序排列最小的等价字符串。
\n\n
示例 1：
\n\n\n输入：s1 = \"parker\", s2 = \"morris\", baseStr = \"parser\"\n输出：\"makkek\"\n解释：根据\n\nA和B中的等价信息，我们可以将这些字符分为[m,p],[a,o],[k,r,s],[e,i]共 4 组。每组中的字符都是等价的，并按字典序排列。所以答案是\"makkek\"。\n
示例 2：
\n\n\n输入：s1 = \"hello\", s2 = \"world\", baseStr = \"hold\"\n输出：\"hdld\"\n解释：根据\n\nA和B中的等价信息，我们可以将这些字符分为[h,w],[d,e,o],[l,r]共 3 组。所以只有 S 中的第二个字符'o'变成'd'，最后答案为\"hdld\"。\n
示例 3：
\n\n\n输入：s1 = \"leetcode\", s2 = \"programs\", baseStr = \"sourcecode\"\n输出：\"aauaaaaada\"\n解释：我们可以把\n\nA和B中的等价字符分为[a,o,e,r,s,c],[l,p],[g,t]和[d,m]共 4 组，因此S中除了'u'和'd'之外的所有字母都转化成了'a'，最后答案为\"aauaaaaada\"。\n
\n\n
提示：
\n\n1 <= s1.length, s2.length, baseStr <= 1000s1.length == s2.lengths1, s2, and baseStr 仅由从 'a' 到 'z' 的小写英文字母组成。\\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>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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 \\u5df2\\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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"book": null,
"isSubscribed": false,
"isDailyQuestion": false,
"dailyRecordStatus": null,
"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "\"parker\"\n\"morris\"\n\"parser\"\n\"hello\"\n\"world\"\n\"hold\"\n\"leetcode\"\n\"programs\"\n\"sourcecode\"",
"__typename": "QuestionNode"
}
}
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