LeetCode 1061. Lexicographically Smallest Equivalent String

原题链接在这里：https://leetcode.com/problems/lexicographically-smallest-equivalent-string/

题目：

Given strings A and B of the same length, we say A[i] and B[i] are equivalent characters. For example, if A = "abc" and B = "cde", then we have 'a' == 'c', 'b' == 'd', '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 A and B above, S = "eed", "acd", and "aab" are equivalent strings, and "aab" is the lexicographically smallest equivalent string of S.

Return the lexicographically smallest equivalent string of S by using the equivalency information from A and B.

Example 1:

Input: A = "parker", B = "morris", S = "parser"
Output: "makkek"
Explanation: Based on the equivalency information in A and B, 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: A = "hello", B = "world", S = "hold"
Output: "hdld"
Explanation:  Based on the equivalency information in A and B, we can group their characters as [h,w], [d,e,o], [l,r]. So only the second letter 'o' in S is changed to 'd', the answer is "hdld".

Example 3:

Input: A = "leetcode", B = "programs", S = "sourcecode"
Output: "aauaaaaada"
Explanation:  We group the equivalent characters in A and B as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in S except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".

Note:

1. String A, B and S consist of only lowercase English letters from 'a'- 'z'.
2. The lengths of string A, B and S are between 1 and 1000.
3. String A and B are of the same length.

题解：

A and B are equal, for each index, the corresponding character in A and B should be in the same union.

When do the union, union by rank. a<c, a is c's parent.

Later, for each character of S, find its ancestor and append it to result.

Time Complexity: O((m+n)logm). m = A.length(), n = S.length(). find takes O(logm).

With path compression and union by rank, amatorize O(1).

Space: O(m).

AC Java:

 class Solution {
     Map<Character, Character> parent = new HashMap<>();

     public String smallestEquivalentString(String A, String B, String S) {
         for(int i = 0; i<A.length(); i++){
             char a = A.charAt(i);
             char b = B.charAt(i);

             if(find(a) != find(b)){
                 union(a, b);
             }
         }

         StringBuilder sb = new StringBuilder();
         for(int i = 0; i<S.length(); i++){
             char anc = find(S.charAt(i));
             sb.append(anc);
         }

         return sb.toString();
     }

     private char find(char c){
         parent.putIfAbsent(c, c);
         if(c != parent.get(c)){
             char anc = find(parent.get(c));
             parent.put(c, anc);
         }

         return parent.get(c);
     }

     private void union(char a, char b){
         char c1 = find(a);
         char c2 = find(b);
         if(c1 < c2){
             parent.put(c2, c1);
         }else{
             parent.put(c1, c2);
         }
     }
 }