# 997. 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:

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`

{% hint style="info" %}
可以看作信任的關係是一個有向圖，建立一個計算in/out degree的counter array，看看每個人的indegree - outdegree是否為N - 1。
{% endhint %}

```cpp
// Directed graph with counting degrees
int findJudge(int N, vector<vector<int>>& trust) { // time: O(T + N); space: O(N)
    vector<int> degree(N + 1, 0);
    for (vector<int>& t : trust) {
        --degree[t[0]], ++degree[t[1]];
    }
    for (int i = 1; i <= N; ++i) {
        if (degree[i] == N - 1) return i;
    }
    return -1;
}
```

{% content-ref url="277.-find-the-celebrity" %}
[277.-find-the-celebrity](277.-find-the-celebrity)
{% endcontent-ref %}
