The Celebrity Problem

Moderate
0/80
Average time to solve is 30m
profile
Contributed by
202 upvotes
Asked in companies
OlaVisaGoogle

Problem statement

There are ‘N’ people at a party. Each person has been assigned a unique id between 0 to 'N' - 1(both inclusive). A celebrity is a person who is known to everyone but does not know anyone at the party.

Given a helper function ‘knows(A, B)’, It will returns "true" if the person having id ‘A’ know the person having id ‘B’ in the party, "false" otherwise. Your task is to find out the celebrity at the party. Print the id of the celebrity, if there is no celebrity at the party then print -1.

Note: 
1. The helper function ‘knows’ is already implemented for you.
2. ‘knows(A, B)’ returns "false", if A doesn't know B.
3. You should not implement helper function ‘knows’, or speculate about its implementation.
4. You should minimize the number of calls to function ‘knows(A, B)’.
5. There are at least 2 people at the party.
6. At most one celebrity will exist.
Detailed explanation ( Input/output format, Notes, Images )
Input format: 
The first line of input contains an integer ‘T’ denoting the number of test cases. The description of  ‘T’ test cases follows.

The first line of each test case contains an integer ‘N’, representing the number of people in the party.
Output format : 
For each test case, print an integer representing the id of the celebrity. If there is no celebrity at the party then print -1.
Note: 
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 50
2 <= N <= 10^4

Where ‘T’ is the total number of test cases, ‘N’ is the number of people at the party.

Time Limit: 1sec
Sample Input 1:
1
2
Call function ‘knows(0, 1)’ // returns false
Call function ‘knows(1, 0)’ // returns true
Sample Output 1:
0
Explanation For Sample Input 1:
In the first test case, there are 2 people at the party. When we call function knows(0,1), it returns false. That means the person having id ‘0’ does not know a person having id ‘1'. Similarly, the person having id ‘1’  knows a person having id ‘0’ as knows(1,0) returns true. Thus a person having id ‘0’ is a celebrity because he is known to everyone at the party but doesn't know anyone.
Sample Input 2:
1
2
Call ‘knows(0, 1)’ // returns true
Call ‘knows(1, 0)’ // returns true
2
Call ‘knows(0, 1)’ // returns false
Call ‘knows(1, 0)’ // returns false
Sample Output 2:
-1
-1
Explanation For Sample Input 2:
In first test case, there are 2 people at the party. The person having id ‘0’  knows a person having id ‘1’. The person having id ‘1’  knows a person having id ‘0’. Thus there is no celebrity at the party, because both know each other. 
In second test case, there are 2 people at the party. The person having id ‘0’ does not knows a person having id ‘1’. The person having id ‘1’  also does not knows a person having id ‘0’. Thus there is no celebrity at the party, because both does not know each other.
Hint

Can you model the problem as a graph problem by considering each person as a node in the graph?

Approaches (4)
Graph

This problem can be modelled as a graph problem. Consider a directed graph having ‘N’ nodes numbered from 0 to ‘N’ - 1.  If the helper function ‘knows(i, j)’ returns true, then it means that there is a directed edge from node ‘i’ to node ‘j’.  We can observe that if the celebrity is present then it is represented by a global sink i.e node that has indegree n-1 and outdegree 0.

 

  1. Make two integer arrays ‘INDEGREE’ and  ‘OUTDEGREE’ of size ‘N’. And fill both of them by 0. These arrays will represent the indegree and outdegree of each node.
  2. Run a nested loop where the outer loop ‘i’ ranges from 0 to ‘N’ - 1 and the inner loop ‘j’ ranges from 0 to ‘N’ - 1,  and for each pair (i, j) if ‘knows(i, j)’ return true, then increment ‘OUTDEGREE[i]’ by 1 and ‘INDEGREE[j]’ by 1.
  3. Initialize an integer variable ‘CELEBRITY' = -1.
  4. Run a loop where ‘i’ ranges from 0 to ‘N’ - 1, and find ‘i’ for which ‘INDEGREE[i]’ is ‘N’ - 1 and ‘OUTDEGREE[i]’ is 0 if such ‘i’ exist then assign ‘CELEBRITY’:= ‘i’, otherwise keep the value of ‘CELEBRITY’ as -1.
  5. Return ‘CELEBRITY’.
Time Complexity

O(N*N), where ‘N’ is the number of people at the party.

 

Because, the nested loop will take the time of the order of N*N.

Space Complexity

O(N), where ‘N’ is the number of people at the party.

 

The size of the array ‘INDEGREE’ and ‘OUTDEGREE’  will be of the order of ‘N’.

Video Solution
Unlock at level 3
(75% EXP penalty)
Code Solution
(100% EXP penalty)
The Celebrity Problem
Full screen
Console