Find the Town Judge - codepath/compsci_guides GitHub Wiki

Problem Highlights

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?
  • Does the graph have to a connected graph?
    • Yes, this graph does not need to be a connected graph. There would be no town judge if graph is disconnected.
  • How many people could there be in town?
    • Between 1 and 1000 people
  • Are all trust pairs unique, there are no duplicate pairs correct?
    • Yes all trust pairs are unique and there are no duplicate pairs
  • What is the time and space constraints for this problem?
    • Let's go with O(V + E) time and O(V) space.
HAPPY CASE
Input: n = 2, trust = [1,2](/codepath/compsci_guides/wiki/1,2)
Output: 2

Input: n = 3, trust = [1,3],[2,3](/codepath/compsci_guides/wiki/1,3],[2,3)
Output: 3

EDGE CASE
Input: n = 1, trust = []
Output: 1

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Graph Problems, common solution patterns include:

  • DFS/BFS: This may not a connected graph, a DFS/BFS approach will be less optimal in identifying the town judge.
  • Adjacency List: We already have an adjacency list known as trust that we can use.
  • Adjacency Matrix: We can use an adjacency matrix to store the graph, but this will make the problem more complicated
  • Topological Sort: In order to have a topological sorting, the graph must not contain any cycles. We cannot apply this sort to this problem because we can have cycles in our graph.
  • Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Since all trust pairs are unique, we can ensure that any person with n-1 trust is the town judge, by reducing trust for the truster and increase trust for the trustee. This way a person who trust another person cannot achieve n-1 trust.

1. Create a list of of person and their trust value
2. Increase trust value of trustee and decrease trust value of truster for each pair
3. Check if anyone achieves n-1 trust. This person is the town judge
4. If no one achieves n-1 trust, then there is no town judge.

⚠️ Common Mistakes

  • Generating a union set here, would work. However, this will look like a brute force approach to this problem.

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        # Create a list of of person and their trust value
        numTrust = [0] * (n + 1)

        # Increase trust value of trustee and decrease trust value of truster for each pair
        for person1, person2 in trust:
            numTrust[person1] -= 1
            numTrust[person2] += 1
        
        # Check if anyone achieves n-1 trust. This person is the town judge
        for i in range(1, len(numTrust)):
            if numTrust[i] == n - 1:
                return i
        
        # If no one achieves n-1 trust, then there is no town judge
        return -1

class Solution {
    public int findJudge(int n, int[][] trust) {
        if (trust.length == 0 && n == 1) 
            return 1;
        
        // Create a list of of person and their trust value
        int[] count = new int[n + 1];

        // Increase trust value of trustee and decrease trust value of truster for each pair
        for (int[] person : trust) {
            count[person[0]]--;
            count[person[1]]++;
        }

        // Check if anyone achieves n-1 trust. This person is the town judge
        for (int person = 0; person < count.length; person++) {
            if (count[person] == n - 1) return person;
        }

        // If no one achieves n-1 trust, then there is no town judge
        return -1;
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

  • Trace through your code with an input to check for the expected output
  • Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume V represents the number of vertices/nodes. Assume E represents the number of edges

  • Time Complexity: O(V+E) We will need to visit each vertex/nodes and their edges.
  • Space Complexity: O(V) accounting for the use of a array to track the number of trust.