1061. Lexicographically Smallest Equivalent String.cpp

/*
Problem link: https://leetcode.com/problems/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.

*/


// solved by Milon

class Solution {
public:

    char find(const char &ch) {
        if (par.find(ch) == par.end() || par[ch] == ch) {
            return ch;
        }
        return par[ch] = find(par[ch]);
    }

    string smallestEquivalentString(const string &s1, const string &s2, string baseStr) {
        int len = s1.size();
        for (int i = 0; i < len; i++) {
            char p1 = find(s1[i]);
            char p2 = find(s2[i]);
            if (p1 < p2)
                par[p2] = p1;
            else
                par[p1] = p2;
        }
        for (auto &ch : baseStr) {
            ch = find(ch);
        }
        return baseStr;
    }
private:
    unordered_map<char, char> par;
};