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0997 - Find the Town Judge (Easy)

https://leetcode.com/problems/find-the-town-judge/

Problem Statement​

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-1otherwise.

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 <= 10^4
  • trust[i].length == 2
  • All the pairs of trust are unique.
  • ai != bi
  • 1 <= ai, bi <= n

Approach 1: Counting (Two Arrays)​

Written by @wingkwong
class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
		// indegree (in) = trusted by others
		// outdegree (out) = trust others
        vector<int> in(n + 1), out(n + 1);
		// calculate the in and out
        for (auto x : trust) in[x[1]]++, out[x[0]]++;
		// iterate each person
        for (int i = 1; i <= n; i++) {
			// the judge will be the one with indegree = n - 1 and outdegree = 0
            if (in[i] == n - 1 && out[i] == 0) {
				// found the judge
                return i;
            }
        }
        // cannot identify -> return -1
        return -1;
    }
};

Approach 2: Counting (One Array)​

We can further simplify Approach 1 with one array since we know indegrees - outdegrees = n - 1.

Written by @wingkwong
class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        // n people
        vector<int> v(n + 1);
        for (auto x : trust) {
            // out: trust other
            v[x[0]]--;
            // in: trusted by others
            v[x[1]]++;
        }
        for (int i = 1; i <= n; i++) {
            // we are looking for indegrees - outdegrees = n - 1
            if (v[i] == n - 1) {
                // if so, the i-th person will be the judge
                return i;
            }
        }
        // cannot identify -> return -1
        return -1;
    }
};