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
Note:
1 <= N <= 1000trust.length <= 10000trust[i]are all differenttrust[i][0] != trust[i][1]1 <= trust[i][0], trust[i][1] <= N
Related Topics:
Graph
Similar Questions:
// 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;
}
};