# 1061. Lexicographically Smallest Equivalent String

Medium
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.
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"
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.

### 解題

Runtime: 0 ms, faster than 100%
Memory Usage: 2.1 MB, less than 100%
func smallestEquivalentString(s1 string, s2 string, baseStr string) string {
parents := make([]int, 26) // 儲存parents
for i:=0; i<26; i++ {
parents[i] = i
}
for i:=0; i<len(s1); i++ {
a, b := int(s1[i] - 'a'), int(s2[i] - 'a')
pa, pb := find(a, &parents), find(b, &parents)
if pa < pb {
parents[pb] = pa // 紀錄比較小的當parent
} else if pa > pb {
parents[pa] = pb
}
}
ans := make([]byte, 0)
for i:=0; i<len(baseStr); i++ {
ans = append(ans, byte(find(int(baseStr[i] - 'a'), &parents) + 'a'))
}
return string(ans)
}
func find(x int, parents *[]int) int {
if (*parents)[x] == x { return x }
(*parents)[x] = find((*parents)[x], parents)
return (*parents)[x]
}