# 277 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`.
Complexity: O(N) time, O(1) space
The Idea: To find the candidate we simply iterate through the candidates and swap when the candidate knows someone. The idea is that if A knows B, then A is immediately out of the picture, so moving on to B, we don't have to check if B knows A or any of the previous candidates because that will be taken care of in the second pass. The first pass is just to make an guaranteed elimination of candidates; that is, we simply move forward with what we know as a valid candidate as a celebrity. Once we have our candidate selected we have to make sure that it is a celebrity. A celebrity must follow 3 rules: 1) Knows nobody 2) Everyone knows the celebrity 3) Don't check yourself!
// Forward declaration of the knows API.
bool knows(int a, int b);
class Solution {
public:
int findCelebrity(int n) {
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 && (knows(candidate, i) || !knows(i, candidate))) return -1;
}
return candidate;
}
};