Problem description:

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.

Solution:

The idea is to first find a person that does not know everyone.

Since celebrity don’t know anyone else, so if candidate knows i, candidate is definitely not celebrity.

  1. If A knows B: A must not be celebrity, B possible
  2. If A doesn’t know B: A possible, B must not be celebrity.

After we got a candidate like this, we need to check two things.

  1. This candidate does not know any one else
  2. Everyone knows the candidate

One thing to improve is that we can use the index to help. Since we’ve already did the knows(candidate, i) in previous loop, we only need to check !knows(i, candidate) if i> candidate.

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// Forward declaration of the knows API.
bool knows(int a, int b);

class Solution {
public:
    int findCelebrity(int n) {
        if(n<=1) return n; 

        int candidate = 0; //set first person as candidate

        for(int i=1; i< n; i++){
            if(knows(candidate, i)){ 
                candidate = i;
            }
        }

        for(int j=0; j< n; j++){
            // speed up the searching process, because part of the check did it before.
            if(i<candidate && knows(candidate, i) || !knows(i, candidate)) return -1;
            if(i>candidate && !knows(i, candidate)) return -1;
        }
        return candidate;
    }
};

time complexity: $O(n)$
space complexity: $O(1)$
reference:
https://goo.gl/tyBj1e