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399. Evaluate Division

Difficulty Topics

Description

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.

Solution

evaluate-division.py
class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        cal = {}
        eq = defaultdict(dict)

        for (a, b), v in zip(equations, values):
            eq[a][b] = v
            eq[b][a] = 1.0 / v

        res = []

        def go(a, b):
            if a not in eq and b not in eq:
                return -1.0

            queue = deque([(a, 1.0)])
            visited = set()

            while queue:
                node, curr = queue.popleft()
                visited.add(node)

                for nei, val in eq[node].items():
                    if nei == b:
                        return curr * val
                    if nei in visited:
                        continue

                    queue.append((nei, val * curr))

            return -1.0

        for a, b in queries:
            res.append(go(a, b))

        return res