Problem:

In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

1. The town judge trusts nobody.
2. Everybody (except for the town judge) trusts the town judge.
3. There is exactly one person that satisfies properties 1 and 2.

You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi. If a trust relationship does not exist in trust array, then such a trust relationship does not exist.

Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.

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

Constraints:

• 1 <= n <= 1000
• 0 <= trust.length <= 104
• trust[i].length == 2
• All the pairs of trust are unique.
• ai != bi
• 1 <= ai, bi <= n

Problem Analysis:

1. High-level Strategy:
   • The solution first initializes a list tracker with n elements, all initialized to zero. This list will keep track of the net trust scores for each person.
   • Then, it iterates through the list of trusts. For each pair (a, b) in trust, it decrements the trust score of a-1 and increments the trust score of b-1.
   • Finally, it iterates through the tracker list and checks if any person has a trust score of n-1. If found, it returns the index of that person plus one (to match the 1-based indexing used in the problem statement). If no such person is found, it returns -1.
2. Complexity:
   • Time Complexity:
     • The loop that iterates through the list of trusts runs in O(E), where E is the number of trust relationships.
     • The loop that iterates through the tracker list runs in O(N), where N is the number of people.
     • Thus, the overall time complexity is O(N + E).
   • Space Complexity:
     • The space complexity is O(N) because of the tracker list.

Solutions:

class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        tracker = [0] * n


        for a,b in trust:
            tracker[a-1] -=1
            tracker[b-1] +=1

        # if town judge, trust == n-1, should match
        for i in range(0, n):
            if tracker[i] == n-1:
                return i+1
        return -1

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