# find-the-town-judge


Try it on <a href='https://leetcode.com/problems/find-the-town-judge'>leetcode</a>

## Description
<div class="description">
<div><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>.</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>Example 1:</strong></p>

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

<p><strong>Example 2:</strong></p>

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

<p><strong>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>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    # The key idea is to visulaize this problem as grpah of n people with trust as
    # graph problem
    # According to the problem Town judge
    # 1.trusts nobody (i.e zero out going connections)
    # 2.everybody trusts him (i.e n incoiming connections)
    #
    # Instead of creating and traversing graph for finding a node with n-1             # connections
    #
    #  A counter array of size n indicating n people can be used to count
    #  number of outgoing connections by incrementing incoming connetion and
    #  decrementing the out going connection
    #
    # Time Complexity : O(T) ,T = length of trust array since 1<= n<=1000 nis               negligible
    # Space Complexity: O(T)
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        cntr = [0] * (n)

        for (a, b) in trust:
            cntr[a - 1] -= 1
            cntr[b - 1] += 1

        for i, cnt in enumerate(cntr):
            if cnt == n - 1:
                return i + 1

        return -1

```