LeetCode - Find the Town Judge(997)

Graph - Easy

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.

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

Constraints:

∙ 1 <= n <= 1000
∙ 0 <= trust.length <= 104
∙ trust[i].length == 2
∙ trust[i] are all different
∙ trust[i][0] != trust[i][1]
∙ 1 <= trust[i][0], trust[i][1] <= n

---

그래프답지 못한 솔루션

class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        normal = []
        # town = []
        dic = {}
        
        for i in range(len(trust)):
            if normal.count(trust[i][0]) == 0:
                normal.append(trust[i][0])
                
            if trust[i][1] not in dic:
                dic[trust[i][1]] = 1
            else:
                dic[trust[i][1]] += 1
        
        dic = sorted(list(dic.items()), key = lambda x : x[1], reverse = True)
        
        if len(normal) == n: # 모든 노드가 다 믿음을 당했을 때.
            return -1
        
        if len(trust) == 0 and n == 1:
            return 1
        elif len(trust) == 0 and n != 1:
            return -1
        
        if dic[0][1] == n - 1:
            return dic[0][0]
        else:
            return -1