# 997. Find the Town Judge

## Description

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. If a trust relationship does not exist in trust array, then such a trust relationship does not exist.

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.
• ai != bi
• 1 <= ai, bi <= n

## Solutions

Solution 1: Counting

We create two arrays $cnt1$ and $cnt2$ of length $n + 1$, representing the number of people each person trusts and the number of people who trust each person, respectively.

Next, we traverse the array $trust$, for each item $[a_i, b_i]$, we increment $cnt1[a_i]$ and $cnt2[b_i]$ by $1$.

Finally, we enumerate each person $i$ in the range $[1,..n]$. If $cnt1[i] = 0$ and $cnt2[i] = n - 1$, it means that $i$ is the town judge, and we return $i$. Otherwise, if no such person is found after the traversal, we return $-1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $trust$.

• class Solution {
public int findJudge(int n, int[][] trust) {
int[] cnt1 = new int[n + 1];
int[] cnt2 = new int[n + 1];
for (var t : trust) {
int a = t[0], b = t[1];
++cnt1[a];
++cnt2[b];
}
for (int i = 1; i <= n; ++i) {
if (cnt1[i] == 0 && cnt2[i] == n - 1) {
return i;
}
}
return -1;
}
}

• class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
vector<int> cnt1(n + 1);
vector<int> cnt2(n + 1);
for (auto& t : trust) {
int a = t[0], b = t[1];
++cnt1[a];
++cnt2[b];
}
for (int i = 1; i <= n; ++i) {
if (cnt1[i] == 0 && cnt2[i] == n - 1) {
return i;
}
}
return -1;
}
};

• class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
cnt1 = [0] * (n + 1)
cnt2 = [0] * (n + 1)
for a, b in trust:
cnt1[a] += 1
cnt2[b] += 1
for i in range(1, n + 1):
if cnt1[i] == 0 and cnt2[i] == n - 1:
return i
return -1


• func findJudge(n int, trust [][]int) int {
cnt1 := make([]int, n+1)
cnt2 := make([]int, n+1)
for _, t := range trust {
a, b := t[0], t[1]
cnt1[a]++
cnt2[b]++
}
for i := 1; i <= n; i++ {
if cnt1[i] == 0 && cnt2[i] == n-1 {
return i
}
}
return -1
}

• function findJudge(n: number, trust: number[][]): number {
const cnt1: number[] = new Array(n + 1).fill(0);
const cnt2: number[] = new Array(n + 1).fill(0);
for (const [a, b] of trust) {
++cnt1[a];
++cnt2[b];
}
for (let i = 1; i <= n; ++i) {
if (cnt1[i] === 0 && cnt2[i] === n - 1) {
return i;
}
}
return -1;
}


• impl Solution {
pub fn find_judge(n: i32, trust: Vec<Vec<i32>>) -> i32 {
let mut cnt1 = vec![0; (n + 1) as usize];
let mut cnt2 = vec![0; (n + 1) as usize];

for t in trust.iter() {
let a = t[0] as usize;
let b = t[1] as usize;
cnt1[a] += 1;
cnt2[b] += 1;
}

for i in 1..=n as usize {
if cnt1[i] == 0 && cnt2[i] == (n as usize) - 1 {
return i as i32;
}
}

-1
}
}