[LeetCode] 399. Evaluate Division_Medium tag: DFS

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.

 

Ideas: T: O(m * n) m = len(quations), n = len(queries)  S: O(m)

note: 

equations: [["a","b"],["b","c"],["bc","cd"]]
values: [2.0,3.0, 3.0]
queries: [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"], ["ba","ad"], ["b","d"]]

should return [6.00000,0.50000,-1.00000,1.00000,-1.00000,-1.00000,-1.00000]

 

1. helper function ： 将equation 和values转换为graph

2. dfs function: 从start -> target 找path，然后将edge 相乘，如果不能找到，返回-1

3. for loop, each query append ans

 

 

Code:

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = self.generateGraph(equations, values)
        ans = []
        for d_e, div in queries:
            if div not in graph or d_e not in graph:
                res = -1.0
            elif div == d_e:
                res = 1
            else:
                visited = set()
                res = self.dfs(d_e, div, 1, visited, graph)
            ans.append(res)
        return ans
    
    def generateGraph(self, equations, values):
        graph = collections.defaultdict(collections.defaultdict)
        for (d_e, di), val in zip(equations, values):
            graph[d_e][di] = val
            graph[di][d_e] = 1/val
        return graph
    
    def dfs(self, cur, targ, cur_res, visited, graph):
        visited.add(cur)
        res = -1.0
        neigs = graph[cur]
        if targ in neigs:
            res = cur_res * neigs[targ]
        else:
            for neig, val in neigs.items():
                if neig in visited:
                    continue
                res = self.dfs(neig, targ, cur_res * val, visited, graph)
                if res != -1:
                    break
        return res