How to Find the Town Judge using Graph Algorithm?

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 <= 1000
trust.length <= 10000
trust[i] are all different
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N

C++ Implementation to Find Town Judge using Graph Algorithm

Let’s use two arrays to count the number of incoming and outgoing edges (trust). For outgoing edges from A to B, we immediately invalidate A being the town judge according to the problem statement.

For incoming edges A to B, we increment the counter count[B]. When we know there are N-1 incoming connections, we make a note on the B. However, if there is a second candidate that has received N-1 connections, we know there are no answers which we can simply return -1.

One corner case is when there is only 1 people and the trust array is empty, the people himself/herself is the judge although he trusts nobody.

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        vector<int> trust_count(N, 0);
        vector<bool> valid(N, true);
        int c = 0, x = -1;
        for (const auto &n: trust) {
            trust_count[n[1] - 1] ++;   
            if (trust_count[n[1] - 1] == N - 1) { // receive N-1 incoming edges
                c ++;
                x = n[1];
                if (c >= 2) return -1; // more than 1 satisfied, thus no judges
            }
            valid[n[0] - 1] = false; // the judge trusts nobody
        }
        if (c == 0) return N == 1 ? 1 : -1; // corner case
        if (valid[x - 1]) return x; // c = 1, also judge trusts nobody.
        return -1;
    }
};

The above C++ implementation runs at O(N) time and space complexity.

–EOF (The Ultimate Computing & Technology Blog) —