# Leetcode 997 - Find the Town Judge with Code

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Here is a step-by-step guide on how to solve the LeetCode problem #997: Find the Town Judge.

### Understanding the Problem

In 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`.

To solve this problem, we need to find the person who is trusted by everyone else, but trusts no one. Here is a plan to solve this problem:

• Create two lists, `in_degree` and `out_degree`, of size `N + 1`. These lists will store the in-degree (number of people trusting a person) and out-degree (number of people a person trusts) of each person. Initialize both lists with all zeros.
• Iterate through the `trust` array and update the in-degree and out-degree of each person. For example, if `trust[i] = [a, b]`, we will increment `in_degree[b]` and `out_degree[a]`.
• Iterate through the `in_degree` and `out_degree` lists. If a person has `in_degree[i] = N - 1` (everyone trusts them) and `out_degree[i] = 0` (they trust no one), then they are the town judge. Return their label `i`.
• If no such person is found, return `-1`.

Now that we have a plan, let’s implement the solution in Python:

``````def findJudge(N: int, trust: List[List[int]]) -> int:
# Create two lists to store the in-degree and out-degree of each person
in_degree = [0] * (N + 1)
out_degree = [0] * (N + 1)

# Update the in-degree and out-degree of each person
for a, b in trust:
in_degree[b] += 1
out_degree[a] += 1

# Find the person with in-degree N - 1 and out-degree 0
for i in range(1, N + 1):
if in_degree[i] == N - 1 and out_degree[i] == 0:
return i

# If no such person is found, return -1
return -1
``````

This solution has a time complexity of O(N) and a space complexity of O(N).

Here are some test cases that you can use to test the solution.

``````# Test case 1
# There are 2 people and person 1 trusts person 2
assert findJudge(2, [[1,2]]) == 2

# Test case 2
# There are 3 people and person 1 trusts person 3, person 2 trusts person 3
assert findJudge(3, [[1,3],[2,3]]) == 3

# Test case 3
# There are 3 people and person 1 trusts person 3, person 2 trusts person 3, person 3 trusts person 1
assert findJudge(3, [[1,3],[2,3],[3,1]]) == -1

# Test case 4
# There is only one person
assert findJudge(1, []) == 1

# Test case 5
# There are 2 people and no trusts are given
assert findJudge(2, []) == -1
``````

I hope these test cases are helpful.

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