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Aroup Goldar Dhruba

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# LeetCode: Find the Town Judge

### Problem Statement

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`

### Solution

``````class Solution {
public:
int findJudge(int N, vector<vector<int>>& trust) {
unordered_map<int,int>trustedBy, trusted;

int judge = -1;
for(int i=1;i<=N;i++)
{
trustedBy[i]=0;
trusted[i]=0;
}
for(int i=0;i<trust.size();i++)
{
trustedBy[trust[i][1]]++;
trusted[trust[i][0]]++;
}
for(int i=1;i<=N;i++)
{
if(trustedBy[i] == N-1 && trusted[i] == 0)
{
judge = i;
break;
}
}
return judge;
}
};
``````

### Solution Thought Process

• First we declare two `unordered_maps` - `trustedBy` and `trusted`
• If a trusts b, then `trustedBy[b] = 1` because b is trusted by a, and `trusted[a] = 1` because a trusted b. For each entry in `trust` we populate `trustedBy` and `trusted`.
• Then for every people, we see if their `trustedBy` is `N-1` and their `trusted` is 0, meaning that they have been trusted by all and they didn't trust anyone. If we found anyone like this, we break the loop and set the people as our judge.

### Complexity

Time Complexity: O(n).

Space Complexity: O(n).