lexicographically_smallest_equivalent_string.dart

/*

-* 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 = "leet-code", s2 = "programs", baseStr = "source-code"
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.

*/

class UnionFind {
  List<int> parent = List.filled(26, 0);

  UnionFind() {
    for (int i = 0; i < 26; ++i) {
      parent[i] = i;
    }
  }

  int find(int a) {
    if (a != parent[a]) {
      parent[a] = find(parent[a]);
    }
    return parent[a];
  }

  void union(int a, int b) {
    int parentA = find(a);
    int parentB = find(b);

    if (parentA == parentB) return;

    if (parentA < parentB) {
      parent[parentB] = parentA;
    } else {
      parent[parentA] = parentB;
    }
  }
}

class A {
  String smallestEquivalentString(String s1, String s2, String baseStr) {
    UnionFind uf = UnionFind();
    for (int i = 0; i < s1.length; ++i) {
      uf.union(s1.codeUnitAt(i) - 'a'.codeUnitAt(0),
          s2.codeUnitAt(i) - 'a'.codeUnitAt(0));
    }

    StringBuffer sb = StringBuffer();
    for (String ch in baseStr.split("")) {
      sb.writeCharCode(
          ('a'.codeUnitAt(0) + uf.find(ch.codeUnitAt(0) - 'a'.codeUnitAt(0))));
    }
    return sb.toString();
  }
}