277. Find the Celebrity
- Total Accepted: 15126
- Total Submissions: 42620
- Difficulty: Medium
- Contributors: Admin
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.
Code
# 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
"""
# pick up two candidates, exclude at least one each time
candidates = set(range(n))
while len(candidates) > 1:
a, b = candidates.pop(), candidates.pop()
if knows(a, b) and not knows(b, a):
candidates.add(b)
elif not knows(a, b) and knows(b, a):
candidates.add(a)
if len(candidates) == 1:
c = candidates.pop()
# when only one candidate is left, we still need to ensure
# his celebrity by definition
for i in xrange(n):
if i != c and not (knows(i, c) and not knows(c, i)):
return -1
return c
else:
return -1
Idea
This is my first attack. Create a set of candidates. Exclude at least one candidate by determining two candidates’ relationship each time. When there is only one candidate c left, we still need to check by definition whether knows(i, c) and not knows(c, i) for all i != c.
An improved idea is to think this problem as numerical comparisons. knows(a,b) means a < b. The celebrity will be the largest element among all peers.
Therefore, the code could be:
# 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
"""
if n == 0:
return -1
x = 0
for i in xrange(1, n):
if knows(x, i):
x = i
for i in xrange(n):
if i != x and not (knows(i, x) and not knows(x, i)):
return -1
return x
Reference: https://discuss.leetcode.com/topic/25720/java-python-o-n-calls-o-1-space-easy-to-understand-solution