Find the town judge

Time: O(T+N); Space: O(N); 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: 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 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

Notes:

  • 1 <= N <= 1000

  • trust.length <= 10000

  • trust[i] are all different

  • trust[i][0] != trust[i][1]

  • 1 <= trust[i][0], trust[i][1] <= N

[1]:

class Solution1(object):
    def findJudge(self, N, trust):
        """
        :type N: int
        :type trust: List[List[int]]
        :rtype: int
        """
        degrees = [0] * N
        for i, j in trust:
            degrees[i-1] -= 1
            degrees[j-1] += 1
        for i in range(len(degrees)):
            if degrees[i] == N - 1:
                return i + 1
        return -1

[2]:

s = Solution1()
N = 2
trust = [[1,2]]
assert s.findJudge(N, trust) == 2
N = 3
trust = [[1,3],[2,3]]
assert s.findJudge(N, trust) == 3
N = 3
trust = [[1,3],[2,3],[3,1]]
assert s.findJudge(N, trust) == -1
N = 3
trust = [[1,2],[2,3]]
assert s.findJudge(N, trust) == -1
N = 4
trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
assert s.findJudge(N, trust) == 3