277. Find the Celebrity

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.

Problem URL

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Solution

party上有n个人，名人的定义是聚会上的其他人都知道他\她但是他\她其余人都不认识。找到名人，或者返回-1；

According to the definition, if people in this party have a relationship grahp, the celebrity is a node only has indegree but does not have outdegree. So we could assign celebrity to 0 first, using a for loop, if celebrity knows I, let I become celebrity.

If there is a celebrity in this party, the first for loop could find him/her definitely. Beacause if celebrity is fake, the guys behind him is not celebrity because there is one guy who does not know him. So the real celebrity is ahead of him. There are contradictory.

After that, use another for loop to validate celebrity. If he/she knows somebody or somebody do not know he/her, return -1. #not include himself in if judgement.

Code

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

Time Complexity: O(n)
Space Complexity: O(1)

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