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 - 1people 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.

Firstly, we make sure that there is only one or no celebrity. And if we get true when call function knows with two people A and B, it indicates A knows B, according to the celebrity’s definition, A must be not a celebrity since a celebrity doesn’t know anybody. And B maybe a celebrity. Otherwise, we get false from the function call, A doesn’t know B, which means B can’t be a celebrity because all the other people should all know a celebrity, and A maybe a celebrity.

And another detail to figure out, if there is only one or none people, there is no celebrity.

So we can find one candidate through one iteration. At the beginning we choose the first people as candidate, then call function knows with candidate and every people, then update the candidate after every checking. At last, we check the candidates with the rules of celebrity to verify he is or not a real celebrity, it not return -1.

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/* 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) {
if (n < 2) return -1;
int candidate = 0;
for (int i = 1; i < n; i++) {
if (knows(candidate, i)) candidate = i;
}

for (int i = 0; i < n; i++) {
if (i != candidate) {
if (knows(candidate, i) || !knows(i, candidate)) return -1;
}
}
return candidate;
}
}

The complexity: O(n) / O(1)