"""
990 Satisfiability of Equality Equations

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 '='
"""
class Solution:
    # my union find solution
    def equationsPossible(self, equations):
        """
        :type equations: list[str]
        :rtype: bool
        """
        root = list(range(26))
        def find(x):
            """
            union find
            """
            while root[x] != x:
                x = root[x]            
            return x
        
        # parse equations
        equal, not_equal = [], []
        for e in equations:
            a, b = ord(e[0]) - ord('a'), ord(e[3]) - ord('a')
            if e[1] == '=':
                equal.append((a, b))
            else:
                not_equal.append((a, b))
        
        # union those equals together
        for a, b in equal:
            ra, rb = find(a), find(b)
            root[ra] = rb
        
        # now check those non-equals
        for a, b in not_equal:
            ra, rb = find(a), find(b)
            if ra == rb:
                return False
        
        return True

equations = ["a==b","b!=c","c==a"]
print(Solution().equationsPossible(equations))