<p>You are given an array of variable pairs <code>equations</code> and an array of real numbers <code>values</code>, where <code>equations[i] = [A<sub>i</sub>, B<sub>i</sub>]</code> and <code>values[i]</code> represent the equation <code>A<sub>i</sub> / B<sub>i</sub> = values[i]</code>. Each <code>A<sub>i</sub></code> or <code>B<sub>i</sub></code> is a string that represents a single variable.</p>

<p>You are also given some <code>queries</code>, where <code>queries[j] = [C<sub>j</sub>, D<sub>j</sub>]</code> represents the <code>j<sup>th</sup></code> query where you must find the answer for <code>C<sub>j</sub> / D<sub>j</sub> = ?</code>.</p>

<p>Return <em>the answers to all queries</em>. If a single answer cannot be determined, return <code>-1.0</code>.</p>

<p><strong>Note:</strong> 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.</p>

<p><strong>Note:&nbsp;</strong>The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;c&quot;]], values = [2.0,3.0], queries = [[&quot;a&quot;,&quot;c&quot;],[&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;e&quot;],[&quot;a&quot;,&quot;a&quot;],[&quot;x&quot;,&quot;x&quot;]]
<strong>Output:</strong> [6.00000,0.50000,-1.00000,1.00000,-1.00000]
<strong>Explanation:</strong> 
Given: <em>a / b = 2.0</em>, <em>b / c = 3.0</em>
queries are: <em>a / c = ?</em>, <em>b / a = ?</em>, <em>a / e = ?</em>, <em>a / a = ?</em>, <em>x / x = ? </em>
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
note: x is undefined =&gt; -1.0</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;c&quot;],[&quot;bc&quot;,&quot;cd&quot;]], values = [1.5,2.5,5.0], queries = [[&quot;a&quot;,&quot;c&quot;],[&quot;c&quot;,&quot;b&quot;],[&quot;bc&quot;,&quot;cd&quot;],[&quot;cd&quot;,&quot;bc&quot;]]
<strong>Output:</strong> [3.75000,0.40000,5.00000,0.20000]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;]], values = [0.5], queries = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;c&quot;],[&quot;x&quot;,&quot;y&quot;]]
<strong>Output:</strong> [0.50000,2.00000,-1.00000,-1.00000]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= equations.length &lt;= 20</code></li>
	<li><code>equations[i].length == 2</code></li>
	<li><code>1 &lt;= A<sub>i</sub>.length, B<sub>i</sub>.length &lt;= 5</code></li>
	<li><code>values.length == equations.length</code></li>
	<li><code>0.0 &lt; values[i] &lt;= 20.0</code></li>
	<li><code>1 &lt;= queries.length &lt;= 20</code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>1 &lt;= C<sub>j</sub>.length, D<sub>j</sub>.length &lt;= 5</code></li>
	<li><code>A<sub>i</sub>, B<sub>i</sub>, C<sub>j</sub>, D<sub>j</sub></code> consist of lower case English letters and digits.</li>
</ul>

class Solution {
public:
    double findCost(string src, string dest, map<string, vector<pair<string, double>>>& adj){
        map<string, bool> vis;
        map<string, double> cost;

        queue<string> q;
        q.push(src);
        cost[src]= 1.0;
        vis[src]= 1;

        while(!q.empty()){
            string frontVar= q.front();
            q.pop();

            for(auto child: adj[frontVar]){
                if(vis[child.first])    continue;

                cost[child.first]= cost[frontVar] * child.second;

                if(child.first == dest) return cost[dest];

                q.push(child.first);
                vis[child.first]= 1;
            }
        }

        return (cost[dest] != 0.00) ? cost[dest] : -1.00;
    }

    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        
        map<string, vector<pair<string, double>>> adj;
        map<string, bool> isPresent;
        vector<double> result;
        map<pair<string, string>, double> cost;

        int i= 0;

        for(auto equation: equations){
            string var1= equation[0];
            string var2= equation[1];

            isPresent[var1]= 1;
            isPresent[var2]= 1;

            cost[make_pair(var1, var1)]= 1;
            cost[make_pair(var2, var2)]= 1;

            adj[var1].push_back(make_pair(var2, values[i]));
            adj[var2].push_back(make_pair(var1, 1/values[i]));

            i++;
        }

        for(auto query: queries){
            string var1= query[0];
            string var2= query[1];

            if(!isPresent[var1] || !isPresent[var2]){
                result.push_back(-1.0);
                continue;
            }

            result.push_back(findCost(query[0], query[1], adj));
        }

        return result;
    }
};
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