[LeetCode#277] Find the Celebrity

Problem:

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 - 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 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.

Analysis:

Even though the problem is not hard, the efficiency could vary a lot, and the concisness of the code could also vary a lot!!!
I have explored following mistakes and implementation until I reach the final solution.

Mistake 1:
Put return in "if-else". Note this is a very common mistake, you should always guanratee a default return at last.

        if (candidates.size() != 1) {
            return -1;
        } else{
            for (int i : candidates)
                return i;
        }

Line 27: error: missing return statement

Mistake 2:  Modify sets when iterate on it.
public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n is invalid");
        HashSet<Integer> candidates = new HashSet<Integer> ();
        for (int i = 0; i < n; i++)
            candidates.add(i);
        for (int i = 0; i < n; i++) {
            for (int j : candidates) {
                if (i != j) {
                    if (knows(i, j))
                        candidates.remove(i);
                    else
                        candidates.remove(j);
                }
            }
        }
        if (candidates.size() == 1) {
            for (int i : candidates)
                return i;
        }
        return -1;
    }
}

Runtime Error Message:
Line 26: java.util.ConcurrentModificationException

Note: this is absolutely allowed!!! A way to avoid this is to use a HashSet to record all the elements you want to remove. When you finish the iteration, then you can remove those elements through "set.removeAll()" method to remove them. 

public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n is invalid");
        HashSet<Integer> candidates = new HashSet<Integer> ();
        HashSet<Integer> excludes = new HashSet<Integer> ();
        for (int i = 0; i < n; i++)
            candidates.add(i);
        for (int i = 0; i < n; i++) {
            for (int j : candidates) {
                if (i != j) {
                    if (knows(i, j)) {
                        excludes.add(i);
                    } else{
                        excludes.add(j);
                    }
                }
            }
            candidates.removeAll(excludes);
        }
        if (candidates.size() == 1) {
            for (int i : candidates)
                return i;
        }
        return -1;
    }
}

Input:
0 knows 1; 1 knows 0.
Output:
1
Expected:
-1

As you can see from the error notification, the above solution could still be wrong. The problem is that even there maybe only "1 element" left in the candidates set, it may not be the answer.

case 1: iff such celebrity exist, there must one element left.
case 2: iff not exist, there also could be elements left in the Set, thus we have to do the final check to resure the left element is a celebrity/ 

Update:
public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n is invalid");
        HashSet<Integer> candidates = new HashSet<Integer> ();
        HashSet<Integer> excludes = new HashSet<Integer> ();
        for (int i = 0; i < n; i++)
            candidates.add(i);
        for (int i = 0; i < n; i++) {
            for (int j : candidates) {
                if (i != j) {
                    if (knows(i, j)) {
                        excludes.add(i);
                    } else{
                        excludes.add(j);
                    }
                }
            }
            candidates.removeAll(excludes);
        }
        for (int i : candidates) {
            for (int j = 0; j < n; j++) {
                if (i != j) {
                    if (knows(i, j) || !knows(j, i))
                        return -1;
                }
            }
        }
        for (int i : candidates)
            return i;
        return -1;
    }
}

We have used two HashSets in the above solution, and for each non-repeative comparision, we could eliminate one candidate out. Thus for the above code the time complexity is actually O(n).

Actually, using above conclusion, we could write the code into a more elegant way. We could use a stack to guarantee:
1. if there is a celebrity, it must remain in the stack(as long as there is an element)
2. there is no repeatitive comparision among candidate. 

We know for each comparision, we must be able to eliminate one element out.
1. iff A knows B, A must not the celebrity, since celebrity knows no one.
2. iff A did not know B, B must not the celebrity, since everyone knows celebrity.

while (stack.size() > 1) {
    int i = stack.pop();
    int j = stack.pop();
    if (knows(i, j))
        stack.push(j);
    else
        stack.push(i);
    }
}

Solution:

public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n is invalid");
        if (n == 1)
            return 0;
        Stack<Integer> stack = new Stack<Integer> ();
        for (int i  = 0; i < n; i++)
            stack.push(i);
        while (stack.size() > 1) {
            int i = stack.pop();
            int j = stack.pop();
            if (knows(i, j))
                stack.push(j);
            else
                stack.push(i);
        }
        int j = stack.pop();
        for (int i = 0; i < n; i++) {
            if (i != j) {
                if (!knows(i, j) || knows(j, i))
                    return -1;
            }
        }
        return j;
    }
}