# The Celebrity Problem

Posted: 19 Dec, 2020
Difficulty: Moderate

## PROBLEM STATEMENT

#### 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.
``````
##### 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
`````` Approach 1

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’.