Leetcode: Find the Town Judge

Find the Town Judge



Similar Problems:


In a town, there are N people labelled 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 trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge. Otherwise, return -1.

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

Example 4:

Input: N = 3, trust = [[1,2],[2,3]]
Output: -1

Example 5:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3

Note:

  1. 1 <= N <= 1000
  2. trust.length <= 10000
  3. trust[i] are all different
  4. trust[i][0] != trust[i][1]
  5. 1 <= trust[i][0], trust[i][1] <= N

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


// Blog link: https://code.dennyzhang.com/find-the-town-judge
// Basic Ideas: array
// l[i]: how many people trust me. If I trust anyone, I'm not the candidate. Mark it as -1
// Complexity: Time O(n), Space O(n)
func findJudge(N int, trust [][]int) int {
    l := make([]int, N+1)
    for _, t := range trust {
        t1, t2 := t[0], t[1]
        l[t1] = -1
        if l[t2] != -1 {
            l[t2]++
        }
    }

    for i:=1; i<=N; i++ {
        if l[i] == N-1 {
            return i
        }
    }
    return -1
}
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