Soyo

May Leetcode Challenge - Day 10

Find the Town Judge

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:

The town judge trusts nobody.
Everybody (except for the town judge) trusts the town judge.
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 <= N <= 1000
  • trust.length <= 10000
  • trust[i] are all different
  • trust[i][0] != trust[i][1]
  • 1 <= trust[i][0], trust[i][1] <= N

Solution

Just a straight loop to count number of trust for each person and how many person they trust

class Solution {
    func findJudge(_ N: Int, _ trust: [[Int]]) -> Int {
        var trustMe = Array(repeating: 0, count: N + 1)
        var iTrustSomeone = Array(repeating: 0, count: N + 1)
        
        for t in trust {
            trustMe[t[1]] += 1
            iTrustSomeone[t[0]] += 1
        }
        
        for i in 1...N {
            if (trustMe[i] == N - 1) && (iTrustSomeone[i] == 0) {
                return i
            }
        }
        
        return -1
        
    }
}

We could optimize memory by using only trustMe array

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