Evaluate division

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:

        graph = defaultdict(defaultdict)

        def backtrack_evaluate(curr_node, target_node, acc_product, visited):
            visited.add(curr_node)
            ret = -1.0
            neighbors = graph[curr_node]
            if target_node in neighbors:
                ret = acc_product * neighbors[target_node]
            else:
                for neighbor, value in neighbors.items():
                    if neighbor in visited:
                        continue
                    ret = backtrack_evaluate(
                        neighbor, target_node, acc_product * value, visited)
                    if ret != -1.0:
                        break
            visited.remove(curr_node)
            return ret

        # Step 1). build the graph from the equations
        for (dividend, divisor), value in zip(equations, values):
            # add nodes and two edges into the graph
            graph[dividend][divisor] = value
            graph[divisor][dividend] = 1 / value

        # Step 2). Evaluate each query via backtracking (DFS)
        #  by verifying if there exists a path from dividend to divisor
        results = []
        for dividend, divisor in queries:
            if dividend not in graph or divisor not in graph:
                # case 1): either node does not exist
                ret = -1.0
            elif dividend == divisor:
                # case 2): origin and destination are the same node
                ret = 1.0
            else:
                visited = set()
                ret = backtrack_evaluate(dividend, divisor, 1, visited)
            results.append(ret)

        return results

Evaluate Division

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [A_i, B_i] and values[i] represent the equation A_i / B_i = values[i]. Each A_i or B_i is a string that represents a single variable.

You are also given some queries, where queries[j] = [C_j, D_j] represents the j^th query where you must find the answer for C_j / D_j = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Note: The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.

Example 1:

Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
Explanation: 
Given: a / b = 2.0, b / c = 3.0
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? 
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
note: x is undefined => -1.0

Example 2:

Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
Output: [3.75000,0.40000,5.00000,0.20000]

Example 3:

Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
Output: [0.50000,2.00000,-1.00000,-1.00000]

Constraints:

1 <= equations.length <= 20
equations[i].length == 2
1 <= A_i.length, B_i.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= C_j.length, D_j.length <= 5
A_i, B_i, C_j, D_j consist of lower case English letters and digits.