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Commit 535dcc6

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Add medium problem.
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‎Leetcode/medium/evaluate-division.py‎

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"""
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# EVALUATE DIVISION
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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.
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You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.
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Return the answers to all queries. If a single answer cannot be determined, return -1.0.
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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.
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Example 1:
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Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
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Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
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Explanation:
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Given: a / b = 2.0, b / c = 3.0
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queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
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return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
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Example 2:
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Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
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Output: [3.75000,0.40000,5.00000,0.20000]
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Example 3:
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Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
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Output: [0.50000,2.00000,-1.00000,-1.00000]
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Constraints:
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1 <= equations.length <= 20
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equations[i].length == 2
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1 <= Ai.length, Bi.length <= 5
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values.length == equations.length
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0.0 < values[i] <= 20.0
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1 <= queries.length <= 20
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queries[i].length == 2
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1 <= Cj.length, Dj.length <= 5
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Ai, Bi, Cj, Dj consist of lower case English letters and digits.
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"""
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class Solution:
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def calcEquation(self, equations, values, queries):
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graph = {}
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for i in range(len(equations)):
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num, den = equations[i]
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ans = values[i]
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if num not in graph:
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graph[num] = {num: 1}
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graph[num][den] = ans
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if den not in graph:
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graph[den] = {den: 1}
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graph[den][num] = (1 / ans)
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results = []
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for n, d in queries:
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visited = set()
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visited.add(n)
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results.append(self.dfs(n, d, 1, visited, graph))
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return results
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def dfs(self, num, den, value, visited, graph):
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if num not in graph:
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return -1.0
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possibleValues = graph[num]
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if den in possibleValues:
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return value * possibleValues[den]
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for item in possibleValues:
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if item in visited:
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continue
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visited.add(item)
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check = self.dfs(item, den, value * possibleValues[item], visited, graph)
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if check != -1.0:
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return check
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visited.remove(item)
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return -1.0

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