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 the celebrity, but the celebrity does not know any of them.
Now you want to find out who the celebrity is or verify that there is not one.
You are only allowed to ask questions like: "Hi, A. Do you know B?" to get information about 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) that tells you whether a knows b.
Implement a function int findCelebrity(n). There will be exactly one celebrity if they are at the party.
Return the celebrity's label if there is a celebrity at the party. If there is no celebrity, return -1.

Example 1:

Input: graph = [[1,1,0],[0,1,0],[1,1,1]]
Output: 1
Explanation: There are three persons labeled with 0, 1 and 2. graph[i][j] = 1 means person i knows person j, otherwise graph[i][j] = 0 means person i does not know person j.
The celebrity is the person labeled as 1 because both 0 and 2 know him but 1 does not know anybody.

Example 2:

Input: graph = [[1,0,1],[1,1,0],[0,1,1]]
Output: -1
Explanation: There is no celebrity.

Constraints:

n == graph.length == graph[i].length
2 <= n <= 100
graph[i][j] is 0 or 1.
graph[i][i] == 1

Follow up: If the maximum number of allowed calls to the API knows is 3 * n, could you find a solution without exceeding the maximum number of calls?

Approach 1: Brute Force

As per the problem statement, for a given person i, we can check whether or not i is a celebrity by using the knows(...) API to see if everybody knows i, and that i knows nobody.
Therefore, the simplest way of solving this problem is to go through each of the people in turn, and check whether or not they are a celebrity.
Algorithm It's best to define a separate isCelebrity(...) function that takes the ID number of a specific person and returns true if they are a celebrity and false if not. This avoids the need for complex loop-break conditions, thus keeping the code cleaner.
One edge case we need to be cautious of is not asking person i if they know themselves.
This can be handled by a check for i == j at the start of the main loop of isCelebrity(...) and then simply continue-ing when it is true.


public class Solution extends Relation {
   private int numberOfPeople;
   public int findCelebrity(int n) {
      numberOfPeople = n;
      for (int i = 0; i < n; i++) {
            if (isCelebrity(i)) {
               return i;
            }
      }
      return -1;
   }

   private boolean isCelebrity(int i) {
      for (int j = 0; j < numberOfPeople; j++) {
            if (i == j) continue; // Don't ask if they know themselves.
            if (knows(i, j) || !knows(j, i)) {
               return false;
            }
      }
      return true;
   }
}

Complexity Analysis

We don't know what time and space the knows(...) API uses.
Because it's not our concern, we'll assume it's O(1) for the purpose of analysing our algorithm.
Time Complexity : O(n^2).
Space Complexity : O(1).

277. Find the Celebrity

Find the Celebrity

Approach 1: Brute Force

Complexity Analysis