Find the Celebrity

// 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.

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

public class FindTheCelebrity extends Relation {
    public int findCelebrity(int n) {
        //initialize candidate to 0
        int candidate = 0;
       
        //find viable candidate
        for(int i = 1; i < n; i++) {
            if(knows(candidate, i)) {
                candidate = i;
            }
        }
       
       
        //check that everyone else knows the candidate
        for(int i = 0; i < n; i++) {
            //if the candidate knows the current person or the current person does not know the candidate, return -1 (candidate is not a celebrity)
            if(i != candidate && knows(candidate, i) || !knows(i, candidate)) {
                return -1;
            }
        }
       
        //return the celebrity
        return candidate;
    }
}