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
Algorithm
We can solve this problem by visualizing the graph. Draw N vertices's and start drawing the relationships.
[1,3] means vertex 1 trust vertex 3.
[2,3] means vertex 2 trust vertex 3.
Now create two arrays, indegree and outdegree.
We need to check:
if(indegree[x]==N-1 && outdegree[i]==0){
return x;
}
i.e if the number of edges incoming to the vertex is N-1 (which means N-1 person are trusting x), but there is no outgoing edge (x is not trusting anyone), then that edge represents a 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
Algorithm
We can solve this problem by visualizing the graph. Draw N vertices's and start drawing the relationships.
[1,3] means vertex 1 trust vertex 3.
[2,3] means vertex 2 trust vertex 3.
Now create two arrays, indegree and outdegree.
- indegree[x] represents the numbers of edges incoming to a vertex x.
- outdegree[x] represents the numbers of edges outgoing from vertex x.
We need to check:
if(indegree[x]==N-1 && outdegree[i]==0){
return x;
}
i.e if the number of edges incoming to the vertex is N-1 (which means N-1 person are trusting x), but there is no outgoing edge (x is not trusting anyone), then that edge represents a judge.
Java Solution
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