# Leetcode online Find the judge, graph with highest degree

Here is the problem.

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] != trust[i]
• 1 <= trust[i], trust[i] <= N

And here is my solution.

(Logic :- Check the degree of every node, +1 for incoming and -1 for outgoing. If any node is having degree as N-1 then that is the node.)

public int findJudge(int N, int[][] trust) {
// Create graph of N and then check degree, should be N-1
final int NOT_FOUND = -1;
final int trustArrayLength = trust.length;

// Degree array should have value starting from 1 to N+1
final int[] degreeArray = new int[N + 1];

for (int i = 0; i < trustArrayLength; i++) {
int[] itemInTrustArray = trust[i];

// Since its outbound connection, decrease the degree by 1.
degreeArray[itemInTrustArray]--;

// Since its inbound connection, increase the degree by 1.
degreeArray[itemInTrustArray]++;
}

// Now iterate though the degreeArray to find the index having degree as N-1.
for (int i = 1; i <= N; i++) {
if (degreeArray[i] == N - 1) {
return i;
}
}
return NOT_FOUND;
}


Pleae let me know, the area of improvement.

This is a wonderfully succinct solution. I will make one point:

final int[] degreeArray = new int[N + 1];


This creates a never-used int at degreeArray. I understand that this was a choice so as to be able to use a simple access by value of the trustees:

degreeArray[itemInTrustArray]--;


In the interest of creating the minimum number of objects necessary, and thus using the least memory possible, I would recommend initializing degreeArray to length N

final int[] degreeArray = new int[N];


And then left shifting your insert by value statements

degreeArray[itemInTrustArray--]--;
degreeArray[itemInTrustArray--]++;


and finally updating your final for loop to account for this change to the zero-based indexing inherent to arrays

// Now iterate though the degreeArray to find the index having degree as N-1.
for (int i = 0; i < N; i++) {


Since you are working with int primitives, the math operators here would add only a near-vanishing amount to overall runtime, if that is a concern.

• Thank you, it was really helpful. Apr 6, 2019 at 16:33
• Great!! solution as I wondering why everyone is initializing it as N+1 Apr 11, 2020 at 14:15
• Great!! but this doesn't work, can you please share full code Apr 11, 2020 at 14:30