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[1/14] 1061. Lexicographically Smallest Equivalent String

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작성일 2023.01.14 17:49

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1061. Lexicographically Smallest Equivalent String

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.

For example, if 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:

Reflexivity: 'a' == 'a'.
Symmetry: 'a' == 'b' implies 'b' == 'a'.
Transitivity: '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.

Example 1:

Input: s1 = "parker", s2 = "morris", baseStr = "parser"
Output: "makkek"
Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i].
The characters in each group are equivalent and sorted in lexicographical order.
So the answer is "makkek".

Example 2:

Input: s1 = "hello", s2 = "world", baseStr = "hold"
Output: "hdld"
Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r].
So only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld".

Example 3:

Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"
Output: "aauaaaaada"
Explanation: 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".

Constraints:

1 <= s1.length, s2.length, baseStr <= 1000
s1.length == s2.length
s1, s2, and baseStr consist of lowercase English letters.

Accepted

61.5K

Submissions

79.9K

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링크

https://leetcode.com/problems/lexicographically-smallest-equivalent-string/ 1801 회 연결

익명
작성일 2023.01.14 17:50

import collections
class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:

        lex_dic = collections.defaultdict(set)
        LEN = len(s1)
        
        for i in range(LEN):
            lex_dic[s1[i]].add(s2[i])
            lex_dic[s2[i]].add(s1[i])
        res = []
        
        
        memo = {}
        def dfs(char, visited):
            if char in memo: return memo[char]
            visited.add(char)
            res = char
            for ch in lex_dic[char]:
                if ch not in visited: 
                    res = min(res, dfs(ch, visited))
            
            return res
                
        
        for char in baseStr:
            visited = set()
            min_char = dfs(char, visited)
            res.append(min_char)
            for c in visited: memo[c] = min_char
        return "".join(res)