# 399. 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> equations, vector& values, vector> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector.

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.

O(M+N)
O(E)

## Code

``````class Solution {
Map<String, Map<String, Double>> g = new HashMap<>();
public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
for(int i = 0; i < equations.size(); i++){ // construct graph
List<String> equation = equations.get(i);
String x = equation.get(0);
String y = equation.get(1);
double value = values[i];
if(!g.containsKey(x)){
g.put(x, new HashMap<>());
}
g.get(x).put(y, value);

if(!g.containsKey(y)){
g.put(y, new HashMap<>());
}
g.get(y).put(x, 1.0/value);
}

double[] res = new double[queries.size()];
for(int i = 0; i < queries.size();i++){
res[i] = divide(queries.get(i).get(0), queries.get(i).get(1), new HashSet<String>());
}
return res;
}

private double divide(String x, String y, Set<String> visited){
if(!g.containsKey(x)) return -1;
if(x.equals(y)) return 1;
if(g.get(x).containsKey(y)) return g.get(x).get(y);