1061.lexicographically-smallest-equivalent-string.py

from lc import *

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        p = {}
        def find(x):
            return x if x==p.get(x,x) else find(p[x])
        def union(a,b):
            x,y = find(a), find(b)
            p[max(x,y)] = min(x,y)
        for a,b in zip(s1,s2):
            union(a,b)
        return ''.join(map(find,baseStr))

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        return (p:={},f:=lambda x:x if x==p.get(x,x) else f(p[x]),[p:=p|{max(w:=[*map(f,v)]):min(w)} for v in zip(s1,s2)]) and ''.join(map(f,baseStr))


test('''

1061. Lexicographically Smallest Equivalent String
Medium

925

67

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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.

''')