997

997. Find the Town Judge (Easy)

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.

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

Companies:
Amazon, Google

Related Topics:
Array, Hash Table, Graph

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Solution 1.

// OJ: https://leetcode.com/problems/find-the-town-judge/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        vector<int> indegree(N + 1), outdegree(N + 1);
        for (auto &t : trust) {
            outdegree[t[0]]++;
            indegree[t[1]]++;
        }
        int judge = -1;
        for (int i = 1; i <= N; ++i) {
            if (indegree[i] != N - 1 || outdegree[i] != 0) continue;
            if (judge != -1) return false;
            judge = i;
        }
        return judge;
    }
};