399. Evaluate Division

Medium

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

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.

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.

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 ]

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 <= Ai.length, Bi.length <= 5

• values.length == equations.length

• 0.0 < values[i] <= 20.0

• 1 <= queries.length <= 20

• queries[i].length == 2

• 1 <= Cj.length, Dj.length <= 5

• Ai, Bi, Cj, Dj consist of lower case English letters and digits.

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        self.d = defaultdict(list)
        for index, [A,B] in enumerate(equations):
            self.d[A].append([B, values[index]])
            self.d[B].append([A, 1/ values[index]])
        result = []
        for query in queries:
            result.append(self.bfs(query))
        return result
        
    def bfs(self, query):
        start, end = query
        if start not in self.d or end not in self.d:
            return -1.0
        dq = deque([(start, 1.0)])
        visited = set()
        while dq:
            node, val = dq.popleft()
            if node == end:
                return val
            visited.add(node)
            for adj, value in self.d[node]:
                if adj not in visited:
                    dq.append((adj, val*value))
        return -1.0