Find the Celebrity

Time: O(N); Space: O(1); medium

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 knows(a, b) -> bool which tells you whether A knows B. Implement a function findCelebrity(n) -> int.

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.

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.

Notes:

  • The directed graph is represented as an adjacency matrix, which is an n x n matrix where a[i][j] = 1 means person i knows person j while a[i][j] = 0 means the contrary.

  • Remember that you won’t have direct access to the adjacency matrix.

[ ]:

# def knows(a, b):
#     """
#   :type a: int - person a
#   :type b: int - person b
#     """
#   return a boolean, whether a knows b

[1]:

class Solution1(object):
    def knows(self, a, b, cache):
        cacheKey = (a,b)

        if cacheKey not in cache:
            cache[cacheKey] = knows(a,b)

        return cache[cacheKey]

    def findCelebrity(self, n):
        """
        :type n: int
        :rtype: int
        """
        candidate = 0
        cache = dict()

        # Find the candidate
        for person in range(1, n):
            if self.knows(candidate, person, cache):    # All candidates < i are not celebrity candidates.
                candidate = person

        # Verify the candidate
        for person in range(n):
            if person != candidate and (self.knows(candidate, person, cache) or not self.knows(person, candidate, cache)):
                return -1

        return candidate

[ ]:

class Solution2(object):
    def findCelebrity(self, n):
        """
        :type n: int
        :rtype: int
        """
        candidate = 0
        cache = dict()

        for person in range(1,n):
            if self.knows(candidate, person, cache):
                candidate = person

        if all( ( self.knows(person, candidate, cache) and not self.knows(candidate, person, cache) )
                for person in range(0,n) if person != candidate ):
            return candidate
        else:
            return -1

[ ]:

# s = Solution1()
# n =
# assert s.findCelebrity(n) ==