Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.
Example:
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 ].
The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.
According to the example above:
equations = [ ["a", "b"], ["b", "c"] ], values = [2.0, 3.0], queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.
class Solution { public: vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) { unordered_map<string, unordered_map<string, double>> hash; for(int i = 0; i < equations.size(); i++) { hash[equations[i].first][equations[i].second] = values[i]; hash[equations[i].second][equations[i].first] = 1/values[i]; } for(auto val: hash) hash[val.first][val.first] = 1; for(auto val1: hash) { for(auto val2: hash) for(auto val3: hash) if(hash[val1.first].count(val3.first) && hash[val2.first].count(val1.first)) hash[val2.first][val3.first] = hash[val2.first][val1.first]*hash[val1.first][val3.first]; } vector<double> ans; for(auto val: queries) ans.push_back(hash[val.first].count(val.second)?hash[val.first][val.second]:-1); return ans; } };
