LeetCode [399] Evaluate Division

 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.

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class Solution { Map<String, Map<String, Double>> graph = new HashMap<>(); int n; Set<String> visited = new HashSet<>(); public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) { n = equations.size(); for(int i=0; i<n; ++i){ String u = equations.get(i).get(0), v = equations.get(i).get(1); double r = values[i]; graph.computeIfAbsent(u, (k->new HashMap<>())).put(v, r); graph.computeIfAbsent(v, (k->new HashMap<>())).put(u, 1/r); } double[] ret = new double[queries.size()]; for(int i=0; i<queries.size(); ++i){ visited.clear(); ret[i] = getWeight(queries.get(i).get(0), queries.get(i).get(1)); } return ret; } double getWeight(String start, String end){ if(!graph.containsKey(start) || !graph.containsKey(end)) return -1.0; visited.add(start); if(graph.get(start).containsKey(end)) return graph.get(start).get(end); for(Map.Entry<String, Double> e : graph.get(start).entrySet()){ String next = e.getKey(); if(visited.contains(next)) continue; double t = getWeight(next, end); if(t!=-1.0) return t*e.getValue(); } return -1.0; } }