Find the Town Judge
<p>In a town, there are <code>n</code> people labeled from <code>1</code> to <code>n</code>. There is a rumor that one of these people is secretly the town judge.</p>

<p>If the town judge exists, then:</p>

<ol>
	<li>The town judge trusts nobody.</li>
	<li>Everybody (except for the town judge) trusts the town judge.</li>
	<li>There is exactly one person that satisfies properties <strong>1</strong> and <strong>2</strong>.</li>
</ol>

<p>You are given an array <code>trust</code> where <code>trust[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> representing that the person labeled <code>a<sub>i</sub></code> trusts the person labeled <code>b<sub>i</sub></code>. If a trust relationship does not exist in <code>trust</code> array, then such a trust relationship does not exist.</p>

<p>Return <em>the label of the town judge if the town judge exists and can be identified, or return </em><code>-1</code><em> otherwise</em>.</p>

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

<pre>
<strong>Input:</strong> n = 2, trust = [[1,2]]
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> n = 3, trust = [[1,3],[2,3]]
<strong>Output:</strong> 3
</pre>

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

<pre>
<strong>Input:</strong> n = 3, trust = [[1,3],[2,3],[3,1]]
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= trust.length &lt;= 10<sup>4</sup></code></li>
	<li><code>trust[i].length == 2</code></li>
	<li>All the pairs of <code>trust</code> are <strong>unique</strong>.</li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
</ul>
Graph; Array; Hash Table
class Solution {
    public int findJudge(int n, int[][] trust) {
        int[] cnt1 = new int[n + 1];
        int[] cnt2 = new int[n + 1];
        for (var t : trust) {
            int a = t[0], b = t[1];
            ++cnt1[a];
            ++cnt2[b];
        }
        for (int i = 1; i <= n; ++i) {
            if (cnt1[i] == 0 && cnt2[i] == n - 1) {
                return i;
            }
        }
        return -1;
    }
}