LeetCode 0997. Find the Town Judge Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)