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 cnt1cnt1 and cnt2cnt2 of length n+1n + 1, representing the number of people each person trusts and the number of people who trust each person, respectively.

Next, we traverse the array trusttrust, for each item [ai,bi][a_i, b_i], we increment cnt1[ai]cnt1[a_i] and cnt2[bi]cnt2[b_i] by 11.

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

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

Python Code
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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 Code
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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++ Code
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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 Code
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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 Code
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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 Code
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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
    }
}