Algorithm


Problem Name: 997. Find the Town Judge

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

Code Examples

#1 Code Example with Java Programming

Code - Java Programming


class Solution {
  public int findJudge(int n, int[][] trust) {
    int[] trustScore = new int[n + 1];
    for (int[] trustPair : trust) {
      trustScore[trustPair[1]]++;
      trustScore[trustPair[0]]--;
    }
    for (int i = 1; i  < = n; i++) {
      if (trustScore[i] == n - 1) {
        return i;
      }
    }
    return -1;
  }
}
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Input

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n = 2, trust = [[1,2]]

Output

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2

#2 Code Example with Javascript Programming

Code - Javascript Programming


const findJudge = function(N, trust) {
  const arr = new Array(N + 1).fill(0)
  for(let [t, ted] of trust) {
    arr[t]--
    arr[ted]++
  }
  for(let i = 1; i  < = N; i++) {
    if(arr[i] === N - 1) return i
  }
  return -1
};
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Input

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cmd
n = 2, trust = [[1,2]]

Output

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2

#3 Code Example with Python Programming

Code - Python Programming


class Solution:
    def findJudge(self, N: int, trust: List[List[int]]) -> int:
        j, cnt = collections.Counter(b for a, b in trust).most_common(1)[0] if trust else (N, 0)
        return j if j not in {a for a, b in trust} and cnt == N - 1 else -1
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Input

x
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cmd
n = 3, trust = [[1,3],[2,3]]

Output

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cmd
3

#4 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _0997_FindTheTownJudge
    {
        public int FindJudge(int N, int[][] trust)
        {
            var score = new int[N + 1];
            foreach (var pair in trust)
            {
                score[pair[0]]--;
                score[pair[1]]++;
            }

            for (int i = 1; i  <  N + 1; i++)
            {
                if (score[i] == N - 1)
                    return i;
            }

            return -1;
        }
    }
}
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Input

x
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cmd
n = 3, trust = [[1,3],[2,3]]

Output

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cmd
3
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