Question
Formatted question description: https://leetcode.ca/all/277.html
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).
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.
Note:
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.
@tagarray
Algorithm
Set the candidate res
to 0. The principle is to traverse once
. For the traversed person i, if the candidate res
knows i, set the candidate res
to i. After completing the traversal, check whether the candidate res
is really a celebrity, if it is found not to be a celebrity, return 1, if there is no conflict, return res
.
It can also further reduce
the amount of API calls. The method of finding candidates is the same as above, but it is divided into two sections when verifying.
 First verify everyone in front of the candidate. If the candidate knows anyone, or anyone does not know the candidate, return 1 directly.
 Then verify the people behind the candidate. At this time, you only need to verify whether anyone does not know the candidate.
 Because we have ensured that the candidate will not know anyone before candidate
Code
Java

public class Find_the_Celebrity { public class Solution extends Relation { public int findCelebrity(int n) { if (n <= 1) return 1; int celebrity = 0; // pick candidate for (int i = 0; i < n; i++) { if (celebrity != i && knows(celebrity, i)) { celebrity = i; } } // final check and return, eg: everyone in group knows nobody for (int i = 0; i < n; i++) { if (celebrity != i && !(knows(i, celebrity) && !knows(celebrity, i))) { return 1; } } /* // final check and return, eg: everyone in group knows nobody // less api calling with below block for (int i = 0; i < celebrity; ++i) { if (knows(celebrity, i)  !knows(i, celebrity)) return 1; } for (int i = celebrity + 1; i < n; ++i) { if (!knows(i, celebrity)) return 1; } */ return celebrity; } private boolean knows(int i, int celeb) { // stub return false; } } class Relation{ // stub } }

bool knows(int a, int b); class Solution { public: int findCelebrity(int n) { int res = 0; for (int i = 0; i < n; ++i) { if (knows(res, i)) res = i; } for (int i = 0; i < n; ++i) { if (res != i && (knows(res, i)  !knows(i, res))) return 1; } return res; } };

# The knows API is already defined for you. # @param a, person a # @param b, person b # @return a boolean, whether a knows b # def knows(a, b): class Solution(object): def findCelebrity(self, n): """ :type n: int :rtype: int """ cand = 0 for i in range(1, n): if knows(cand, i): cand = i for i in range(n): if i != cand and knows(cand, i) or not knows(i, cand): return 1 return cand