Find the Celebrity

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