Evaluate Division

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>> m;        vector<double> res;        for (int i = 0; i < values.size(); ++i) {            m[equations[i].first].insert(make_pair(equations[i].second, values[i]));            if (values[i])                m[equations[i].second].insert(make_pair(equations[i].first, 1 / values[i]));        }                for (auto i : queries) {            unordered_set<string> s;            double tmp = check(i.first, i.second, m, s);            if (tmp) res.push_back(tmp);            else res.push_back(-1.0);        }        return res;    }        double check(string up, string down, unordered_map<string, unordered_map<string, double>>& m, unordered_set<string>& s) {        if (m[up].find(down) != m[up].end()) return m[up][down];        for (auto i : m[up]) {            if (s.find(i.first) == s.end()) {                s.insert(i.first);                double tmp = check(i.first, down, m, s);                if (tmp) return tmp * i.second;            }        }        return 0;    }};*///------------ Another method -------------class Solution {public:    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {        unordered_map<string, Node*> m;        vector<double> res;        for(int i = 0; i < values.size(); ++i) {            string s0 = equations[i].first, s1 = equations[i].second;            if (m.count(s0) == 0 && m.count(s1) == 0) { //s0 and s1 are both not in m                m[s0] = new Node(values[i]);                m[s1] = new Node(1.0);                m[s1]->parent = m[s0];            }            else if (m.count(s0) == 0) {    //only s0 not in m                m[s0] = new Node(m[s1]->value * values[i]);                m[s0]->parent = m[s1];            }            else if (m.count(s1) == 0) {    //only s1 not in m                m[s1] = new Node(m[s0]->value / values[i]);                m[s1]->parent = m[s0];            }                else //s0 and s1 are both in m                unionNodes(m[s0], m[s1], values[i], m);        }        for (auto i : queries) {            if (m.count(i.first) == 0 || m.count(i.second) == 0 || findParent(m[i.first]) != findParent(m[i.second]))                res.push_back(-1.0);            else                res.push_back(m[i.first]->value / m[i.second]->value);        }        return res;    }    private:    struct Node {        Node* parent;        double value;        Node(double num) : parent(this), value(num) {}    };        void unionNodes(Node* node0, Node* node1, double num, unordered_map<string, Node*>& m) {        Node* parent0 = findParent(node0), *parent1 = findParent(node1);        double ratio = node1->value * num / node0->value;        for (auto it = m.begin(); it != m.end(); ++it)            if (findParent(it->second) == parent0)                it->second->value *= ratio;        parent1 -> parent = parent0;    }        Node* findParent(Node* node) {        if (node->parent == node) return node;        node->parent = findParent(node->parent);        return node->parent;    }};

9 lines "Floyd–Warshall" in Python

class Solution(object):    def calcEquation(self, equations, values, queries):        """        :type equations: List[List[str]]        :type values: List[float]        :type queries: List[List[str]]        :rtype: List[float]        """        quot = collections.defaultdict(dict)        for (num, den), val in zip(equations, values):            quot[num][num] = quot[den][den] = 1.0            quot[num][den], quot[den][num] = val, 1 / val        for k, i, j in itertools.permutations(quot, 3):            if k in quot[i] and j in quot[k]:                quot[i][j] = quot[i][k] * quot[k][j]        return [quot[num].get(den, -1.0) for num, den in queries]