### LeetCode Challenge #399. Evaluate Division

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.

Note: The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.

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 ]
note: x is undefined => -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.
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
void dfs(unordered_map<string,vector<pair<string,double>>> adj, unordered_map<string,bool> vis, double &ans, double prod, string start, string end)
{
if(vis[start])return;
if(start==end)
{
ans=prod;
return;
}

vis[start]=true;

for(auto x:i)
{
}
return;
}
vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {

unordered_set<string>isAvailable;
vector<double> solution;

for(int i=0; i<equations.size(); i++)
{

isAvailable.insert(equations[i][0]);
isAvailable.insert(equations[i][1]);
}

for(int i=0; i<queries.size(); i++)
{
double ans=-1.0,prod=1;
unordered_map<string,bool> vis;

if(isAvailable.find(queries[i][0])!=isAvailable.end() || isAvailable.find(queries[i][1])!=isAvailable.end())
{