1061.Lexicographically Smallest Equivalent String
===

###### tags: `Medium`,`String`,`Union Find`

[1061. Lexicographically Smallest Equivalent String](https://leetcode.com/problems/lexicographically-smallest-equivalent-string/)

### 題目描述

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`.

### 範例

**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.

### 解答

#### C++
##### 思路:
1. equivalent information由s1,s2定義, 同一個位置char視為一組可互相替換, 不同位置具有傳遞性, 如s1='ab', s2='bc', 則'abc'可互相替換視為一組
2. 題目需找baseStr的Lexicographically Smallest equivalent string, 代表找在baseStr中每一個char同一組最小的char替換

##### 做法:
1. 同組找老大(最小的char)用disjoint set

unordered_map<char,char> root_;
    char findRoot(char x)
    {
        if(x == root_[x]) return x;
        char root = findRoot(root_[x]);
        root_[x] = root;
        return root;
    }
    void connect(char x, char y)
    {
        char x_root = findRoot(x);
        char y_root = findRoot(y);
        if(x_root > y_root)
            root_[x_root] = y_root;
        else if(y_root > x_root)
            root_[y_root] = x_root;
    }
    string smallestEquivalentString(string s1, string s2, string baseStr) {
        for(int i = 0; i < 26; i++)
            root_['a'+i] = 'a'+i;
        for(int i = 0; i < s1.length(); i++)
            connect(s1[i], s2[i]);
        string res;
        for(char c : baseStr)
            res += findRoot(c);
        return res;
    }

> [name=XD] [time= Jan 14, 2023]

Time: $O(n*26)$
Extra Space: $O(26)$

class Solution {
public:
    int parents[26];

    int Find(int x) {
        if (parents[x] == -1) return x;
        return parents[x] = Find(parents[x]);
    }

    void Union(int x, int y) {
        x = Find(x);
        y = Find(y);
        if (x != y) parents[max(x, y)] = min(x, y);
    }

    string smallestEquivalentString(string s1, string s2, string baseStr) {
        memset(parents, -1, sizeof(parents));

        for (int i = 0; i < s1.size(); i++) {
            Union(s1[i] - 'a', s2[i] - 'a');
        }
        for (int i = 0; i < baseStr.size(); i++) {
            baseStr[i] = Find(baseStr[i] - 'a') + 'a';
        }
        return baseStr;
    }
};

> [name=Yen-Chi Chen][time=Sat, Jan 14, 2023]

#### Python

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        d = defaultdict(lambda: -1)

        def find(x):
            if d[x] == -1: return x
            d[x] = find(d[x])
            return d[x]
        
        def union(x, y):
            x, y = find(x), find(y)
            if x != y: d[max(x, y)] = min(x, y)

        for a, b in zip(s1, s2):
            union(a, b)
        ans = []
        for c in baseStr:
            ans.append(find(c))
        return ''.join(ans)

> [name=Yen-Chi Chen][time=Sat, Jan 14, 2023]

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        uf = {}

        def find(x):
            uf.setdefault(x, x)
            if uf[x] != x:
                uf[x] = find(uf[x])

            return uf[x]

        def union(x, y):
            x, y = find(x), find(y)
            if x > y:
                uf[x] = y
            else:
                uf[y] = x


        for c1, c2 in zip(s1, s2):
            union(c1, c2)

        res = [find(c) for c in baseStr]
        return "".join(res)

> [name=Ron Chen][time=Wed, Jan 18, 2023]

### Reference
[disjoint set](https://hackmd.io/@CLKO/rkRVS_o-4?type=view)

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