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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:
 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}
. 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 <= 10^{4}
trust[i].length == 2
 All the pairs of
trust
are unique. a_{i} != b_{i}
1 <= a_{i}, b_{i} <= 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] == n1 { 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 } }