Problem URL: Satisfiability of equality equations

We make an undirected graph in which the nodes are integers (as lower-case letters) and each edge connects integers that are equal. We use the union-find algorithm to determine the connected graphs. We keep track of the pairs (a,b) such that a =! b. If the any such pair are in the same connected graph, then return false, otherwise return true.

class Solution:
    def equationsPossible(self, equations: List[str]) -> bool:
        parent, diff = {}, []

        def find(x):
            if x not in parent:
                return x
            return find(parent[x])

        for s in equations:
            a, b = s[0], s[3]
            if s[1] == '=':
                x, y = find(a), find(b)
                if x != y:
                    parent[y] = x
            else:
                diff.append((a, b))

        return all(find(a) != find(b) for a, b in diff)

Time Complexity: O(n^2)
Space Complexity: O(n)