Leetcode 997 - Find the Town Judge with Code

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.

Plan your Solution

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.

Implement your Solution

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).

Test your Solution

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.

Related:

• find the smallest divisor given a threshold
• two sum solution with code
• kth smallest element in a bst with code
• replace words with solution