**I**n a town, there are N people labelled from 1 to N. There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

The town judge trusts nobody.

Everybody (except for the town judge) trusts the town judge.

There is exactly one person that satisfies properties 1 and 2.

You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge. Otherwise, return -1.

**Example 1**:

Input: N = 2, trust = [[1,2]]

Output: 2

**Example 2**:

Input: N = 3, trust = [[1,3],[2,3]]

Output: 3

**Example 3**:

Input: N = 3, trust = [[1,3],[2,3],[3,1]]

Output: -1

**Example 4**:

Input: N = 3, trust = [[1,2],[2,3]]

Output: -1

**Example 5**:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]

Output: 3

**Note**:

1 <= N <= 1000

trust.length <= 10000

trust[i] are all different

trust[i][0] != trust[i][1]

1 <= trust[i][0], trust[i][1] <= N

__Algorithm__We can solve this problem by visualizing the graph. Draw N vertices's and start drawing the relationships.

[1,3] means vertex 1 trust vertex 3.

[2,3] means vertex 2 trust vertex 3.

Now create two arrays, indegree and outdegree.

- indegree[x] represents the numbers of edges incoming to a vertex x.
- outdegree[x] represents the numbers of edges outgoing from vertex x.

We need to check:

*if(indegree[x]==N-1 && outdegree[i]==0){*

*return x;*

*}*

i.e if the number of edges incoming to the vertex is N-1 (which means N-1 person are trusting x), but there is no outgoing edge (x is not trusting anyone), then that edge represents a judge.

__Java Solution__

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