## # Problem

## # Knows API already defined

```
def knows(a: int, b: int) -> bool:
```

## # Solution

Note that it takes

### # Brute Force

Check if each person is a celebrity.

Complexity:

- time:
- 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

Let's ask this question: does A know B? If the answer is yes, then A is defo not a celebrity; o.w., B is not a celebrity. Thus, we could ask this question `n-1`

times and get a `candidate`

for celebrity.

We then check if everyone else knows the `candidate`

and the `candidate`

knows no one else.

Complexity:

- time:
- space: $O(1)

```
def findCelebrity(self, n):
candidate = 0
for i in range(1,n):
if not knows(i,candidate):
### above line could also be: ###
# 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
```