README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0997.Find%20the%20Town%20Judge/README_EN.md	Graph Array Hash Table

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

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 $trust$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}