Tuesday, September 8, 2015

Leetcode: Find the Celebrity

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.
Understand the problem:
The problem can be transformed as a graph problem. We count the in-degree and out-degree for each person. Then find out the person with in-degree n - 1 and out-degree 0. 

Code (Java):
/* The knows API is defined in the parent class Relation.
      boolean knows(int a, int b); */

public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 1) {
            return -1;
        }
        
        int[] inDegree = new int[n];
        int[] outDegree = new int[n];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i != j && knows(i, j)) {
                    outDegree[i]++;
                    inDegree[j]++;
                }
            }
        }
        
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == n - 1 && outDegree[i] == 0) {
                return i;
            }
        }
        
        return -1;
    }
}

Analysis:
Time O(n^2).
Space O(n).

A O(n) time O(1) Space Solution:
Use two pointers, left starts from 0, and right starts from n - 1.
Check if knows(left, right). If yes, left++. Else, right--.
After the first step, we could find out the potential candidate. 
In the second step, we validate the candidate by iterating through all the person again. Each one must know the candidate while the candidate must not know anyone else. 

Code (Java):
/* The knows API is defined in the parent class Relation.
      boolean knows(int a, int b); */

public class Solution extends Relation {
    public int findCelebrity(int n) {
        if (n <= 1) {
            return -1;
        }
        
        int left = 0;
        int right = n - 1;
        
        // Step 1: Find the potential candidate
        while (left < right) {
            if (knows(left, right)) {
                left++;
            } else {
                right--;
            }
        }
        
        // Step 2: Validate the candidate
        int candidate = right;
        for (int i = 0; i < n; i++) {
            if (i != candidate && (!knows(i, candidate) || knows(candidate, i))) {
                return -1;
            }
        }
        
        return candidate;
    }
}



1 comment:

  1. Great articles and nice content... Thanks for posting this informative blog

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