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Problem

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 = 3.0queries 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.

Sol

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        db = defaultdict(lambda :-1.0)
        gh = defaultdict(list)
        
        for ((a,b),val) in zip(equations, values):
            db[(a,b)] = val
            db[(b,a)] = 1.0/val
            db[(a,a)] = db[(b,b)] = 1.0
            gh[a].append(b)
            gh[b].append(a)
        
        def dfs(root,lhs, acc=1.0, seen=set()):
            if lhs in seen:
                return
            else:
                seen.add(lhs)
                for rhs in gh[lhs]:
                    db[(root,rhs)] = acc*db[(lhs,rhs)]
                    db[(rhs,root)] = 1/(acc*db[(lhs,rhs)])
                    dfs(root,rhs,acc*db[(lhs,rhs)],seen)
        
        [dfs(root,root,1.0,set()) for root in gh.keys()]
        return [db[(lhs,rhs)] for (lhs,rhs) in queries]