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:
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;
}
