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

Solution: Union Find

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua, running time: 12 ms, 7.1 MB
class Solution {
public:
  bool equationsPossible(vector<string>& equations) {        
    iota(begin(parents_), end(parents_), 0);    
    for (const auto& eq : equations)     
      if (eq[1] == '=')
        parents_[find(eq[0])] = find(eq[3]);
    for (const auto& eq : equations)     
      if (eq[1] == '!' && find(eq[0]) == find(eq[3]))        
          return false;
    return true;
  }
private:
  array<int, 128> parents_;
  int find(int x) {
    if (x != parents_[x])
      parents_[x] = find(parents_[x]);
    return parents_[x];
  }  
};

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