# [leetcode]Graph-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<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;
}
};  ```

[leetcode]Graph-399. Evaluate Division

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