277. Find the Celebrity (Medium) Facebook LinkedIn

Suppose you are at a party withnpeople (labeled from0ton - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the othern - 1people 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 functionbool knows(a, b)which tells you whether A knows B. Implement a functionint findCelebrity(n), your function should minimize the number of calls toknows.

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.

Solution 1: Brute Force O(n^2); O(1)

For each people i, check if i is the celebrity. There are 2 conditions.
All the other people know i and i does not know any other people.

    public int findCelebrity(int n) {
        for (int i = 0; i < n; i++) {
            int j = 0;
            for (; j < n; j++) {
            //for (int j = 0; i != j && j < n; j++) { //can't put i != j with j < n together, logic error
                if (i != j) { 
                    if (knows(i, j) || !knows(j, i)){ //2 cases: 1.i knows j 2. j don't know i
                        break;
                    }
                }
            }
            if (j == n) { //if j iterates until n - 1, i is the celebrity
                return i;
            }
        }
        return -1;
    }

Solution 2: Two-pass O(n); O(1)

The first pass is to pick out the candidate. If candidate knows i, then let i be the candidate.
The second pass is to check whether the candidate is the real celebrity.

名人谁都不认识, 所以到它那就卡住, 而且不管怎样所有人都认识名人, 所以必定能找到这个名人

    public int findCelebrity(int n) {
        int candidate = 0;
        for (int i = 1; i < n; i++) {
            if (knows(candidate, i)) { //necessary condition
                candidate = i;
            }
        }

        for (int i = 0; i < n; i++) {
            if (i != candidate) {
                if (knows(candidate, i) || !knows(i, candidate)) {
                    return -1;
                }
            }
        }
        return candidate;
    }