Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.

Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).

You are given a helper function bool knows(a, b) which tells you whether A knows B. Implement a function int findCelebrity(n), your function should minimize the number of calls to knows.

Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return -1

two pass

先遍歷一次陣列，找出可能的候選，p從0開始，如果p認識i，那麼p不符合條件，i符合條件，把 i 賦值給p

再遍歷一次陣列，如果當前值 i 不為p，並且如果i 不認識p或者p認識i，說明之前找到的p不符合條件，返回-1

如果第二次遍歷結束沒有不符合條件，返回p

time: O(N), space: O(1)

/* The knows API is defined in the parent class Relation.
      boolean knows(int a, int b); */

public class Solution extends
 
 Relation {
    public int findCelebrity(int n) {
        int p = 0;
        for(int i = 1; i < n; i++) {
            if(knows(p, i))
                p = i;
        }
        
        for(int i = 0; i < n; i++) {
            if(i != p && (!knows(i, p) || knows(p, i)))
                return -1;
        }
        return p;
    }
}

reference: https://leetcode.com/problems/find-the-celebrity/discuss/71227/Java-Solution.-Two-Pass