# Find the Celebrity

Array

## # Solution

knows(a: int, b: int) -> bool takes $O(n)$ time to check if a person is a celebrity.

### # Brute Force

Check if each person is a celebrity.

Complexity

time: $O(n^2)$
space: $O(1)$

def findCelebrity(self, n):
for i in range(n):
founded = True
for j in range(n):
if i == j:
continue
founded = founded and (knows(j,i) and not knows(i,j))
if founded:
return i
return -1


### # Logical Deduction

Does A know B? If the answer is yes, then A is definitely not a celebrity; otherwise, B is not a celebrity. Thus, we could ask this question n-1 times and get a candidate for celebrity.

We then check if:

1. everyone else knows the candidate
2. the candidate knows no one else

Complexity

time: $O(n)$
space: $O(1)$

def findCelebrity(self, n):
candidate = 0

for i in range(1,n):
if not knows(i,candidate):
# OR: if knows(candidate,i):
candidate = i
# check if candidate is a celebrity
for i in range(n):
if i == candidate:
continue
if knows(candidate, i) or not knows(i, candidate):
return -1
return candidate