Evaluate Division
<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>
Depth-First Search; Breadth-First Search; Union Find; Graph; Array; String; Shortest Path
function calcEquation(equations: string[][], values: number[], queries: string[][]): number[] {
    const g: Record<string, [string, number][]> = {};
    const ans = Array.from({ length: queries.length }, () => -1);

    for (let i = 0; i < equations.length; i++) {
        const [a, b] = equations[i];
        (g[a] ??= []).push([b, values[i]]);
        (g[b] ??= []).push([a, 1 / values[i]]);
    }

    for (let i = 0; i < queries.length; i++) {
        const [c, d] = queries[i];
        const vis = new Set<string>();
        const q: [string, number][] = [[c, 1]];

        if (!g[c] || !g[d]) continue;
        if (c === d) {
            ans[i] = 1;
            continue;
        }

        for (const [current, v] of q) {
            if (vis.has(current)) continue;
            vis.add(current);

            for (const [intermediate, multiplier] of g[current]) {
                if (vis.has(intermediate)) continue;

                if (intermediate === d) {
                    ans[i] = v * multiplier;
                    break;
                }

                q.push([intermediate, v * multiplier]);
            }

            if (ans[i] !== -1) break;
        }
    }

    return ans;
}
