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997. Find the Town Judge

EasyOpen on LeetCode Problem statement

Problem Statement

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:

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 an array trust where trust[i] = [a_i, b_i] representing that the person labeled a_i trusts the person labeled b_i.

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.
a_i != b_i
1 <= a_i, b_i <= n

C++ — alternative 1

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        vector <int> v(n+1, 0); // person => {trusted_by_ct, trusts_someone}
        for(auto e: trust)
            v[e[1]]++, v[e[0]]=INT_MIN;
        for(int i=1; i<v.size(); i++)
            if(v[i]==n-1)
                return i;
        return -1;
    }
};

C++ — alternative 2

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        int p[n];   // potential judges
        for (int i=0; i<n; i++){
            p[i]=0;
        }
        for (int i=0; i<trust.size(); i++){
            p[trust[i][0]-1] = INT_MIN;   // eliminated from consideration
            p[trust[i][1]-1]++;   // trusted by 1 more person
        }
        for (int i=0; i<n; i++){
            if (p[i]==n-1)
                return (i+1);
        }
        return -1;
    }
};