990. Satisfiability of Equality Equations

Question:

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: “a==b” or “a!=b”. Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.

Example 1:

Input: ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.

Example 2:

Input: ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Example 3:

Input: ["a==b","b==c","a==c"]
Output: true

Example 4:

Input: ["a==b","b!=c","c==a"]
Output: false

Example 5:

Input: ["c==c","b==d","x!=z"]
Output: true

Note:

1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] and equations[i][3] are lowercase letters
equations[i][1] is either ‘=’ or ‘!’
equations[i][2] is ‘=’

Solution:

Method 1: Union find set beats 100%

 class Solution {
      private static final char EQ = '=';
      private static final char NOT_EQ = '!';
      private char[] uf;
      public boolean equationsPossible(String[] equations) {
          this.uf = new char[26];
          for(int i = 0; i < 26; i++) uf[i] = (char)(i + 'a');
          for(int i = 0; i < equations.length; i++){
              String eq = equations[i];
              if(eq.charAt(1) == EQ){ // Add to union find set
                  union(eq.charAt(0), eq.charAt(3));
              }
          }
          for(int i = 0; i < equations.length; i++){
              String eq = equations[i];
              if(eq.charAt(1) == NOT_EQ){
                  char a = find(eq.charAt(0));
                  char b = find(eq.charAt(3));
                  if(a == b) return false;
              }
          }
          return true;
      }
      private char find(char c){
          if(c != uf[c - 'a']){
              uf[c - 'a'] = find(uf[c - 'a']);
          }
          return uf[c - 'a'];
      }
      private void union(char a, char b){
          char p = find(a);
          char q = find(b);
          uf[p - 'a'] = q;
      }
  }