399 Evaluate Division

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

class Solution:
    def calcEquation(
        self,
        equations: List[List[str]],
        values: List[float],
        queries: List[List[str]]
    ) -> List[float]:
        
        def dfs(cur: str, dst: str, ans: float) -> float:
            if cur == dst:
                return ans
            for adj, adj_val in graph[cur].items():
                if adj not in seen:
                    seen.add(adj)
                    result = dfs(adj, dst, ans * adj_val)
                    if result != -1:
                        return result
                    seen.remove(adj)
            return -1
        
        graph = defaultdict(dict)
        for (a, b), val in zip(equations, values):
            graph[a][b] = val
            graph[b][a] = 1 / val
        res = []
        seen = set()
        for src, dst in queries:
            res.append(dfs(src, dst, 1) if src in graph else -1)
            seen.clear()
        return res