HàPhan 河

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