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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0997_find_the_town_judge;

// #Easy #Array #Hash_Table #Graph #Data_Structure_II_Day_19_Graph
// #Graph_Theory_I_Day_13_Graph_Theory #2022_03_31_Time_3_ms_(80.64%)_Space_49.9_MB_(88.09%)

/**
 * 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] = [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
**/
public class Solution {
    public int findJudge(int n, int[][] trust) {
        if (trust.length < n - 1) {
            return -1;
        }
        int[] trustScores = new int[n];
        for (int[] t : trust) {
            trustScores[t[1] - 1]++;
            trustScores[t[0] - 1]--;
        }
        for (int i = 0; i < n; i++) {
            if (trustScores[i] == n - 1) {
                return i + 1;
            }
        }
        return -1;
    }
}