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