README_EN.md

997. Find the Town Judge

中文文档

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

Python3

class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        N = 1001
        c1, c2 = [0] * N, [0] * N
        for a, b in trust:
            c1[a] += 1
            c2[b] += 1
        for i in range(1, N):
            if c1[i] == 0 and c2[i] == n - 1:
                return i
        return -1

Java

class Solution {
    public int findJudge(int n, int[][] trust) {
        int N = 1001;
        int[] c1 = new int[N];
        int[] c2 = new int[N];
        for (int[] e : trust) {
            ++c1[e[0]];
            ++c2[e[1]];
        }
        for (int i = 1; i < N; ++i) {
            if (c1[i] == 0 && c2[i] == n - 1) {
                return i;
            }
        }
        return -1;
    }
}

TypeScript

function findJudge(n: number, trust: number[][]): number {
    let candidates = new Array(n).fill(0);
    for (let [a, b] of trust) {
        candidates[a - 1] = -1;
        if (candidates[b - 1] >= 0) {
            candidates[b - 1]++;
        }
    }

    for (let i = 0; i < n; i++) {
        if (candidates[i] == n - 1) {
            return i + 1;
        }
    }
    return -1;
}

C++

class Solution {
public:
    int findJudge(int n, vector<vector<int>>& trust) {
        int N = 1001;
        vector<int> c1(N);
        vector<int> c2(N);
        for (auto& e : trust) {
            ++c1[e[0]];
            ++c2[e[1]];
        }
        for (int i = 1; i < N; ++i) {
            if (c1[i] == 0 && c2[i] == n - 1) return i;
        }
        return -1;
    }
};

Go

func findJudge(n int, trust [][]int) int {
	N := 1001
	c1 := make([]int, N)
	c2 := make([]int, N)
	for _, e := range trust {
		c1[e[0]]++
		c2[e[1]]++
	}
	for i := 1; i < N; i++ {
		if c1[i] == 0 && c2[i] == n-1 {
			return i
		}
	}
	return -1
}

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