find-town-judge

Problem

Find Town Judge

Problem Description

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. 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

This problem can be seen as a directed graph，from a -> b, a represent a trust b, as below example pic. so that we can calculate based on the degrees.

• Use a count array -- count[i] -- the number of people trust people i.

• for (a,b), a trust b,

  • count[a]-1

  • count[b]+1 (increase the number of trust people for b)

• after iterate all trust array, iterate count

  • for people i, if count[i]==N-1, meaning has N - 1 people trust i, return i.

  • if not found any count[i]=N-1

  • return -1 (no judge found)

For example:

Find Town Judge

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(N)

• N - the number N

Code

class Solution {
    public int findJudge(int N, int[][] trust) {
        // count start from 1 to N, [1,N]
        int[] count = new int[N + 1];
        // for (a,b) -- a trust b, when a, count[a]--, count[b]++
        for (int[] t : trust) {
            count[t[0]]--;
            count[t[1]]++;
        }
        // iterate count start from 1, if count[i]==N-1, meaning all people trust i, and i does not trust anyone, return i. judge found. otherwise return -1.
        for (int i = 1; i <= N; i++) {
            if (count[i] == N - 1) return i;
        }
        return -1;
    }