LeetCode 398 Evaluate Division - Medium

398. Evaluate Division — Medium

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Problem

399. Evaluate Division -- Medium

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

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = {}
        def build_graph(equations, values):
            def add_edge(f, t, value):
                if f in graph:
                    graph[f].append((t, value))
                else:
                    graph[f] = [(t, value)]
            for vertices, value in zip(equations, values):
                f, t = vertices
                add_edge(f, t, value)
                add_edge(t, f, 1/value)
        def find_path(query):
            b, e = query
            if b not in graph or e not in graph:
                return -1.0
            q = collections.deque([(b, 1.0)])
            visited = set()
            while q:
                front, cur_product = q.popleft()
                if front == e:
                    return cur_product
                visited.add(front)
                for neighbor, value in graph[front]:
                    if neighbor not in visited:
                        q.append((neighbor, cur_product*value))
            return -1.0
        build_graph(equations, values)
        return [find_path(q) for q in queries]