The Celebrity Problem

Difficulty Level : Hard
Last Updated : 22 Jul, 2021

In a party of N people, only one person is known to everyone. Such a person may be present in the party, if yes, (s)he doesn’t know anyone in the party. We can only ask questions like “does A know B? “. Find the stranger (celebrity) in the minimum number of questions.
We can describe the problem input as an array of numbers/characters representing persons in the party. We also have a hypothetical function HaveAcquaintance(A, B) which returns true if A knows B, false otherwise. How can we solve the problem.

Examples:

Input:
MATRIX = { {0, 0, 1, 0},
           {0, 0, 1, 0},
           {0, 0, 0, 0},
           {0, 0, 1, 0} }
Output:id = 2
Explanation: The person with ID 2 does not 
know anyone but everyone knows him

Input:
MATRIX = { {0, 0, 1, 0},
           {0, 0, 1, 0},
           {0, 1, 0, 0},
           {0, 0, 1, 0} }
Output: No celebrity
Explanation: There is no celebrity.

We measure the complexity in terms of calls made to HaveAcquaintance().

Method 1: This uses Graph to arrive at the particular solution.

Approach:
Model the solution using graphs. Initialize indegree and outdegree of every vertex as 0. If A knows B, draw a directed edge from A to B, increase indegree of B and outdegree of A by 1. Construct all possible edges of the graph for every possible pair [i, j]. There are N_C2 pairs. If a celebrity is present in the party, there will be one sink node in the graph with outdegree of zero and indegree of N-1.

Algorithm:

Create two arrays indegree and outdegree, to store the indegree and outdegree
Run a nested loop, the outer loop from 0 to n and inner loop from 0 to n.
For every pair i, j check if i knows j then increase the outdegree of i and indegree of j
For every pair i, j check if j knows i then increase the outdegree of j and indegree of i
Run a loop from 0 to n and find the id where the indegree is n-1 and outdegree is 0

Implementation:

C++

// C++ program to find celebrity
#include <bits/stdc++.h>
#include <list>
using namespace std;
  
// Max # of persons in the party
#define N 8
  
// Person with 2 is celebrity
bool MATRIX[N][N] = {{0, 0, 1, 0},
                    {0, 0, 1, 0},
                    {0, 0, 0, 0},
                    {0, 0, 1, 0}};
  
bool knows(int a, int b)
{
    return MATRIX[a][b];
}
  
// Returns -1 if celebrity
// is not present. If present,
// returns id (value from 0 to n-1).
int findCelebrity(int n)
{
    //the graph needs not be constructed
    //as the edges can be found by
    //using knows function
     
    //degree array;
    int indegree[n]={0},outdegree[n]={0};
     
    //query for all edges
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            int x = knows(i,j);
             
            //set the degrees
            outdegree[i]+=x;
            indegree[j]+=x;
        }
    }
     
    //find a person with indegree n-1
    //and out degree 0
    for(int i=0; i<n; i++)
    if(indegree[i] == n-1 && outdegree[i] == 0)
        return i;
     
    return -1;
}
  
// Driver code
int main()
{
    int n = 4;
    int id = findCelebrity(n);
    id == -1 ? cout << "No celebrity" :
               cout << "Celebrity ID " << id;
    return 0;
}

Java

// Java program to find celebrity
import java.util.*;
class GFG {
 
  // Max # of persons in the party
  static final int N = 8;
 
  // Person with 2 is celebrity
  static int MATRIX[][] = { { 0, 0, 1, 0 },
                           { 0, 0, 1, 0 },
                           { 0, 0, 0, 0 },
                           { 0, 0, 1, 0 } };
 
  static int knows(int a, int b) { return MATRIX[a][b]; }
 
  // Returns -1 if celebrity
  // is not present. If present,
  // returns id (value from 0 to n-1).
  static int findCelebrity(int n)
  {
 
    // the graph needs not be constructed
    // as the edges can be found by
    // using knows function
 
    // degree array;
    int[] indegree = new int[n];
    int[] outdegree = new int[n];
 
    // query for all edges
    for (int i = 0; i < n; i++)
    {
      for (int j = 0; j < n; j++)
      {
        int x = knows(i, j);
 
        // set the degrees
        outdegree[i] += x;
        indegree[j] += x;
      }
    }
 
    // find a person with indegree n-1
    // and out degree 0
    for (int i = 0; i < n; i++)
      if (indegree[i] == n - 1 && outdegree[i] == 0)
        return i;
 
    return -1;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int n = 4;
    int id = findCelebrity(n);
    if (id == -1)
      System.out.print("No celebrity");
    else
      System.out.print("Celebrity ID " + id);
  }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find celebrity
 
# Max # of persons in the party
N = 8
 
# Person with 2 is celebrity
MATRIX = [ [ 0, 0, 1, 0 ],
           [ 0, 0, 1, 0 ],
           [ 0, 0, 0, 0 ],
           [ 0, 0, 1, 0 ] ]
            
def knows(a, b):
     
  return MATRIX[a][b]
 
def findCelebrity(n):
     
    # The graph needs not be constructed
    # as the edges can be found by
    # using knows function
       
    # degree array;
    indegree = [0 for x in range(n)]
    outdegree = [0 for x in range(n)]
       
    # Query for all edges
    for i in range(n):
        for j in range(n):
            x = knows(i, j)
               
            # Set the degrees
            outdegree[i] += x
            indegree[j] += x
       
    # Find a person with indegree n-1
    # and out degree 0
    for i in range(n):
        if (indegree[i] == n - 1 and
           outdegree[i] == 0):
            return i
             
    return -1
     
# Driver code   
if __name__ == '__main__':
     
    n = 4
    id_ = findCelebrity(n)
     
    if id_ == -1:
        print("No celebrity")
    else:
        print("Celebrity ID", id_)
 
# This code is contributed by UnworthyProgrammer

C#

// C# program to find celebrity
using System;
public class GFG
{
 
  // Max # of persons in the party
  static readonly int N = 8;
 
  // Person with 2 is celebrity
  static int[,] MATRIX = { { 0, 0, 1, 0 },
                           { 0, 0, 1, 0 },
                           { 0, 0, 0, 0 },
                           { 0, 0, 1, 0 } };
 
  static int knows(int a, int b) { return MATRIX[a,b]; }
 
  // Returns -1 if celebrity
  // is not present. If present,
  // returns id (value from 0 to n-1).
  static int findCelebrity(int n)
  {
 
    // the graph needs not be constructed
    // as the edges can be found by
    // using knows function
 
    // degree array;
    int[] indegree = new int[n];
    int[] outdegree = new int[n];
 
    // query for all edges
    for (int i = 0; i < n; i++)
    {
      for (int j = 0; j < n; j++)
      {
        int x = knows(i, j);
 
        // set the degrees
        outdegree[i] += x;
        indegree[j] += x;
      }
    }
 
    // find a person with indegree n-1
    // and out degree 0
    for (int i = 0; i < n; i++)
      if (indegree[i] == n - 1 && outdegree[i] == 0)
        return i;
 
    return -1;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int n = 4;
    int id = findCelebrity(n);
    if (id == -1)
      Console.Write("No celebrity");
    else
      Console.Write("Celebrity ID " + id);
  }
}
 
// This code is contributed by aashish1995

Javascript

<script>
 
// JavaScript program to find celebrity
 
    // Max # of persons in the party
      var N = 8;
 
    // Person with 2 is celebrity
    var MATRIX = [ [ 0, 0, 1, 0 ],
                   [ 0, 0, 1, 0 ],
                   [ 0, 0, 0, 0 ],
                   [ 0, 0, 1, 0 ] ];
 
    function knows(a , b) {
        return MATRIX[a][b];
    }
 
    // Returns -1 if celebrity
    // is not present. If present,
    // returns id (value from 0 to n-1).
    function findCelebrity(n) {
 
        // the graph needs not be constructed
        // as the edges can be found by
        // using knows function
 
        // degree array;
        var indegree = Array(n).fill(0);
        var outdegree = Array(n).fill(0);
 
        // query for all edges
        for (var i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                var x = knows(i, j);
 
                // set the degrees
                outdegree[i] += x;
                indegree[j] += x;
            }
        }
 
        // find a person with indegree n-1
        // and out degree 0
        for (i = 0; i < n; i++)
            if (indegree[i] == n - 1 && outdegree[i] == 0)
                return i;
 
        return -1;
    }
 
    // Driver code
     
        var n = 4;
        var id = findCelebrity(n);
        if (id == -1)
            document.write("No celebrity");
        else
            document.write("Celebrity ID " + id);
 
// This code is contributed by todaysgaurav
 
</script>

Output :

Celebrity ID 2

Complexity Analysis:

Time Complexity: O(n²).
A nested loop is run traversing the array, SO the time complexity is O(n²)
Space Complexity: O(n).
Since extra space of size n is required.

Approach :
The problem can be solved using recursion. Say, if the ‘potential celebrity’ of N-1 persons is known, can the solution to N be found from it? A potential celebrity is one who is the only one left after eliminating n-1 people. n-1 people are eliminated with the following strategy:

If A knows B, then A cannot be a celebrity. But B could be.
Else If B knows A, then B cannot be a celebrity. But A could be.

The above-mentioned approach use Recursion to find the potential celebrity among n persons, recursively calls n-1 persons, till the base case of 0 persons is reached. For 0 persons -1 is returned indicating that there are no possible celebrities since there are 0 people. In the ith stage of recursion, the ith person and (i-1)th person are compared to check if anyone of them knows the other. And using the above logic (in the bullet points) the potential celebrity is returned to the (i+1)th stage.

Once the recursive function returns an id. We check if this id does not know anybody else, but all others know this id. If this is true, then this id will be the celebrity.

Algorithm :

Create a recursive function that takes an integer n.
Check the base case, if the value of n is 0 then return -1.
Call the recursive function and get the ID of the potential celebrity from the first n-1 elements.
If the id is -1 then assign n as the potential celebrity and return the value.
If the potential celebrity of first n-1 elements knows n-1 then return n-1, (0 based indexing)
If the celebrity of first n-1 elements does not know n-1 then return id of celebrity of n-1 elements, (0 based indexing)
Else return -1
Create a wrapper function and check whether the id returned by the function is really the celebrity or not.

Implementation:

C++

// C++ program to find celebrity
#include <bits/stdc++.h>
#include <list>
using namespace std;
 
// Max # of persons in the party
#define N 8
 
// Person with 2 is celebrity
bool MATRIX[N][N] = { { 0, 0, 1, 0 },
                      { 0, 0, 1, 0 },
                      { 0, 0, 0, 0 },
                      { 0, 0, 1, 0 } };
 
bool knows(int a, int b) { return MATRIX[a][b]; }
 
// Returns -1 if a 'potential celebrity'
// is not present. If present,
// returns id (value from 0 to n-1).
int findPotentialCelebrity(int n)
{
    // base case - when n reaches 0 , returns -1
    // since n represents the number of people,
    // 0 people implies no celebrity(= -1)
    if (n == 0)
        return -1;
 
    // find the celebrity with n-1
    // persons
    int id = findPotentialCelebrity(n - 1);
 
    // if there are no celebrities
    if (id == -1)
        return n - 1;
 
    // if the id knows the nth person
    // then the id cannot be a celebrity, but nth person
    // could be one
    else if (knows(id, n - 1)) {
        return n - 1;
    }
    // if the nth person knows the id,
    // then the nth person cannot be a celebrity and the id
    // could be one
    else if (knows(n - 1, id)) {
        return id;
    }
 
    // if there is no celebrity
    return -1;
}
 
// Returns -1 if celebrity
// is not present. If present,
// returns id (value from 0 to n-1).
// a wrapper over findCelebrity
int Celebrity(int n)
{
    // find the celebrity
    int id = findPotentialCelebrity(n);
 
    // check if the celebrity found
    // is really the celebrity
    if (id == -1)
        return id;
    else {
        int c1 = 0, c2 = 0;
 
        // check the id is really the
        // celebrity
        for (int i = 0; i < n; i++)
            if (i != id) {
                c1 += knows(id, i);
                c2 += knows(i, id);
            }
 
        // if the person is known to
        // everyone.
        if (c1 == 0 && c2 == n - 1)
            return id;
 
        return -1;
    }
}
 
// Driver code
int main()
{
    int n = 4;
    int id = Celebrity(n);
    id == -1 ? cout << "No celebrity"
             : cout << "Celebrity ID " << id;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG
{
   
    // Max # of persons in the party
    static int N = 8;
 
    // Person with 2 is celebrity
    static int MATRIX[][] = { { 0, 0, 1, 0 },
                              { 0, 0, 1, 0 },
                              { 0, 0, 0, 0 },
                              { 0, 0, 1, 0 } };
 
    static int knows(int a, int b) { return MATRIX[a][b]; }
 
    // Returns -1 if a 'potential celebrity'
    // is not present. If present,
    // returns id (value from 0 to n-1).
    static int findPotentialCelebrity(int n)
    {
        // base case - when n reaches 0 , returns -1
        // since n represents the number of people,
        // 0 people implies no celebrity(= -1)
        if (n == 0)
            return -1;
 
        // find the celebrity with n-1
        // persons
        int id = findPotentialCelebrity(n - 1);
 
        // if there are no celebrities
        if (id == -1)
            return n - 1;
 
        // if the id knows the nth person
        // then the id cannot be a celebrity, but nth person
        // could be one
        else if (knows(id, n - 1) == 1) {
            return n - 1;
        }
        // if the nth person knows the id,
        // then the nth person cannot be a celebrity and the
        // id could be one
        else if (knows(n - 1, id) == 1) {
            return id;
        }
 
        // if there is no celebrity
        return -1;
    }
 
    // Returns -1 if celebrity
    // is not present. If present,
    // returns id (value from 0 to n-1).
    // a wrapper over findCelebrity
    static int Celebrity(int n)
    {
        // find the celebrity
        int id = findPotentialCelebrity(n);
 
        // check if the celebrity found
        // is really the celebrity
        if (id == -1)
            return id;
        else {
            int c1 = 0, c2 = 0;
 
            // check the id is really the
            // celebrity
            for (int i = 0; i < n; i++)
                if (i != id) {
                    c1 += knows(id, i);
                    c2 += knows(i, id);
                }
 
            // if the person is known to
            // everyone.
            if (c1 == 0 && c2 == n - 1)
                return id;
 
            return -1;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        int id = Celebrity(n);
        if (id == -1) {
            System.out.println("No celebrity");
        }
        else {
            System.out.println("Celebrity ID " + id);
        }
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python3 program to find celebrity
 
# Max # of persons in the party
N = 8
 
# Person with 2 is celebrity
MATRIX = [[0, 0, 1, 0],
           [0, 0, 1, 0],
           [0, 0, 0, 0],
           [0, 0, 1, 0]]
 
 
def knows(a, b):
 
    return MATRIX[a][b]
 
# Returns -1 if a potential celebrity
# is not present. If present,
# returns id (value from 0 to n-1).
 
 
def findPotentialCelebrity(n):
 
    # Base case
    if (n == 0):
        return 0;
 
    # Find the celebrity with n-1
    # persons
    id_ = findPotentialCelebrity(n - 1)
 
    # If there are no celebrities
    if (id_ == -1):
        return n - 1
    # if the id knows the nth person
    # then the id cannot be a celebrity, but nth person
    # could be on
    elif knows(id_, n - 1):
          return n - 1
    # if the id knows the nth person
    # then the id cannot be a celebrity, but nth person
    # could be one
    elif knows(n - 1, id_):
          return id_
    # If there is no celebrity
    return - 1
 
# Returns -1 if celebrity
# is not present. If present,
# returns id (value from 0 to n-1).
# a wrapper over findCelebrity
def Celebrity(n):
 
    # Find the celebrity
    id_=findPotentialCelebrity(n)
 
    # Check if the celebrity found
    # is really the celebrity
    if (id_ == -1):
        return id_
    else:
        c1=0
        c2=0
 
        # Check the id is really the
        # celebrity
        for i in range(n):
            if (i != id_):
                c1 += knows(id_, i)
                c2 += knows(i, id_)
 
        # If the person is known to
        # everyone.
        if (c1 == 0 and c2 == n - 1):
            return id_
 
        return -1
 
# Driver code
if __name__ == '__main__':
 
    n=4
    id_=Celebrity(n)
 
    if id_ == -1:
        print("No celebrity")
    else:
        print("Celebrity ID", id_)
 
# This code is contributed by UnworthyProgrammer

Output :

Celebrity ID 2

Complexity Analysis:

Time Complexity: O(n).
The recursive function is called n times, so the time complexity is O(n).
Space Complexity: O(1).
As no extra space is required.

Approach: There are some observations based on elimination technique (Refer Polya’s How to Solve It book).

If A knows B, then A can’t be a celebrity. Discard A, and B may be celebrity.
If A doesn’t know B, then B can’t be a celebrity. Discard B, and A may be celebrity.
Repeat above two steps till there is only one person.
Ensure the remained person is a celebrity. (What is the need of this step?)

Algorithm:

Create a stack and push all the id’s in the stack.
Run a loop while there are more than 1 element in the stack.
Pop top two element from the stack (represent them as A and B)
If A knows B, then A can’t be a celebrity and push B in stack. Else if A doesn’t know B, then B can’t be a celebrity push A in stack.
Assign the remaining element in the stack as the celebrity.
Run a loop from 0 to n-1 and find the count of persons who knows the celebrity and the number of people whom the celebrity knows. if the count of persons who knows the celebrity is n-1 and the count of people whom the celebrity knows is 0 then return the id of celebrity else return -1.

Implementation:

C++

// C++ program to find celebrity
#include <bits/stdc++.h>
#include <list>
using namespace std;
 
// Max # of persons in the party
#define N 8
 
// Person with 2 is celebrity
bool MATRIX[N][N] = {{0, 0, 1, 0},
                    {0, 0, 1, 0},
                    {0, 0, 0, 0},
                    {0, 0, 1, 0}};
 
bool knows(int a, int b)
{
    return MATRIX[a][b];
}
 
// Returns -1 if celebrity
// is not present. If present,
// returns id (value from 0 to n-1).
int findCelebrity(int n)
{
    // Handle trivial
    // case of size = 2
    stack<int> s;
 
    // Celebrity
    int C;
 
    // Push everybody to stack
    for (int i = 0; i < n; i++)
        s.push(i);
 
    // Extract top 2
  
 
    // Find a potential celebrity
    while (s.size() > 1)
    {   int A = s.top();
        s.pop();
        int B = s.top();
        s.pop();
        if (knows(A, B))
        {
          s.push(B);
        }
        else
        {
          s.push(A);
        }
    }
    // If there are only two people
    // and there is no
    // potential candicate
    if(s.empty())
        return -1;
   
   
    // Potential candidate?
    C = s.top();
    s.pop();
 
    // Check if C is actually
    // a celebrity or not
    for (int i = 0; i < n; i++)
    {
        // If any person doesn't
        // know 'C' or 'C' doesn't
        // know any person, return -1
        if ( (i != C) &&
                (knows(C, i) ||
                 !knows(i, C)) )
            return -1;
    }
 
    return C;
}
 
// Driver code
int main()
{
    int n = 4;
    int id = findCelebrity(n);
    id == -1 ? cout << "No celebrity" :
               cout << "Celebrity ID " << id;
    return 0;
}

Java

// Java program to find celebrity using
// stack data structure
import java.util.Stack;
 
class GFG
{
    // Person with 2 is celebrity
    static int MATRIX[][] = { { 0, 0, 1, 0 },
                            { 0, 0, 1, 0 },
                            { 0, 0, 0, 0 },
                            { 0, 0, 1, 0 } };
 
    // Returns true if a knows
    // b, false otherwise
    static boolean knows(int a, int b)
    {
        boolean res = (MATRIX[a][b] == 1) ?
                                     true :
                                     false;
        return res;
    }
 
    // Returns -1 if celebrity
    // is not present. If present,
    // returns id (value from 0 to n-1).
    static int findCelebrity(int n)
    {
        Stack<Integer> st = new Stack<>();
        int c;
 
        // Step 1 :Push everybody
        // onto stack
        for (int i = 0; i < n; i++)
        {
            st.push(i);
        }
 
        while (st.size() > 1)
        {
            // Step 2 :Pop off top
            // two persons from the
            // stack, discard one
            // person based on return
            // status of knows(A, B).
            int a = st.pop();
            int b = st.pop();
 
            // Step 3 : Push the
            // remained person onto stack.
            if (knows(a, b))
            {
                st.push(b);
            }
 
            else
                st.push(a);
        }
       
        // If there are only two people
        // and there is no
        // potential candicate
        if(st.empty())
            return -1;
 
        c = st.pop();
 
        // Step 5 : Check if the last
        // person is celebrity or not
        for (int i = 0; i < n; i++)
        {
            // If any person doesn't
            //  know 'c' or 'a' doesn't
            // know any person, return -1
            if (i != c && (knows(c, i) ||
                          !knows(i, c)))
                return -1;
        }
        return c;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 4;
        int result = findCelebrity(n);
        if (result == -1)
        {
            System.out.println("No Celebrity");
        }
        else
            System.out.println("Celebrity ID " +
                                        result);
    }
}
 
// This code is contributed
// by Rishabh Mahrsee

Python3

# Python3 program to find celebrity
# using stack data structure
 
# Max # of persons in the party
N = 8
 
# Person with 2 is celebrity
MATRIX = [ [ 0, 0, 1, 0 ],
           [ 0, 0, 1, 0 ],
           [ 0, 0, 0, 0 ],
           [ 0, 0, 1, 0 ] ]
 
def knows(a, b):
     
    return MATRIX[a][b]
 
# Returns -1 if celebrity
# is not present. If present,
# returns id (value from 0 to n-1).
def findCelebrity(n):
     
    # Handle trivial
    # case of size = 2
    s = []
 
    # Push everybody to stack
    for i in range(n):
        s.append(i)
 
    # Extract top 2
    A = s.pop()
    B = s.pop()
 
    # Find a potential celebrity
    while (len(s) > 1):
        if (knows(A, B)):
            A = s.pop()
        else:
            B = s.pop()
             
    # If there are only two people
    # and there is no
    # potential candicate
    if(len(s) == 0):
        return -1
         
    # Potential candidate?
    C = s.pop();
 
    # Last candidate was not
    # examined, it leads one
    # excess comparison (optimize)
    if (knows(C, B)):
        C = B
         
    if (knows(C, A)):
        C = A
 
    # Check if C is actually
    # a celebrity or not
    for i in range(n):
     
        # If any person doesn't
        # know 'a' or 'a' doesn't
        # know any person, return -1
        if ((i != C) and
           (knows(C, i) or
           not(knows(i, C)))):
            return -1
 
    return C
     
# Driver code
if __name__ == '__main__':
     
    n = 4
    id_ = findCelebrity(n)
     
    if id_ == -1:
        print("No celebrity")
    else:
      print("Celebrity ID ", id_)
 
# This code is contributed by UnworthyProgrammer

Output :

Celebrity ID 2

Complexity Analysis:

Time Complexity: O(n).
The total number of comparisons 3(N-1), so the time complexity is O(n).
Space Complexity: O(n).
n extra space is needed to store the stack.

Optimal Approach: The idea is to use two pointers, one from start and one from the end. Assume the start person is A, and the end person is B. If A knows B, then A must not be the celebrity. Else, B must not be the celebrity. At the end of the loop, only one index will be left as a celebrity. Go through each person again and check whether this is the celebrity.
The Two Pointer approach can be used where two pointers can be assigned, one at the start and the other at the end, and the elements can be compared and the search space can be reduced.

Algorithm :

Create two indices i and j, where i = 0 and j = n-1
Run a loop until i is less than j.
Check if i knows j, then i can’t be a celebrity. so increment i, i.e. i++
Else j cannot be a celebrity, so decrement j, i.e. j–
Assign i as the celebrity candidate
Now at last check that whether the candidate is actually a celebrity by re-running a loop from 0 to n-1 and constantly checking that if the candidate knows a person or if there is a candidate who does not know the candidate, then we should return -1. else at the end of the loop, we can be sure that the candidate is actually a celebrity.

Implementation:

Java

// Java program to find celebrity
// in the given Matrix of people
// Code by Sparsh_cbs
 
import java.io.*;
 
class GFG {
    public static void main(String[] args)
    {
        int[][] M = { { 0, 0, 1, 0 },
                      { 0, 0, 1, 0 },
                      { 0, 0, 0, 0 },
                      { 0, 0, 1, 0 } };
 
        int celebIdx = celebrity(M, 4);
 
        if (celebIdx == -1)
            System.out.println("No celebrity found!");
        else {
            System.out.println(
                "0-based celebrity index is : " + celebIdx);
        }
    }
    public static int celebrity(int M[][], int n)
    {
        // This function returns the celebrity
        // index 0-based (if any)
 
        int i = 0, j = n - 1;
        while (i < j) {
            if (M[j][i] == 1) // j knows i
                j--;
            else // i knows j
                i++;
        }
        // i points to our celebrity candidate
        int candidate = i;
 
        // Now, all that is left is to check that whether
        // the candidate is actually a celebrity i.e: he is
        // known by everyone but he knows no one
        for (i = 0; i < n; i++) {
            if (i != candidate) {
                if (M[i][candidate] == 0
                    || M[candidate][i] == 1)
                    return -1;
            }
        }
        // if we reach here this means that the candidate
        // is really a celebrity
        return candidate;
    }
}

Python3

# Python3 code
class Solution:
 
    # Function to find if there is a celebrity in the party or not.
    # return index if celebrity else return -1
    def celebrity(self, M, n):
        # code here
        i = 0
        j = n-1
        candidate = -1
        while(i < j):
            if M[j][i] == 1:
                j -= 1
            else:
                i += 1
 
        candidate = i
        for k in range(n):
            if candidate != k:
                if M[candidate][k] == 1 or M[k][candidate] == 0:
                    return -1
 
        return candidate
 
 
n = 4
m = [[0, 0, 1, 0],
     [0, 0, 1, 0],
     [0, 0, 0, 0],
     [0, 0, 1, 0]]
ob = Solution()
print("Celebrity at index "+str(ob.celebrity(m, n)))

Output :

0-based celebrity index is : 2

Complexity Analysis:

Time Complexity: O(n)
Space Complexity: O(1) No extra space is required.

Write code to find celebrity. Don’t use any data structures like graphs, stack, etc… you have access to N and HaveAcquaintance(int, int) only.
Implement the algorithm using Queues. What is your observation? Compare your solution with Finding Maximum and Minimum in an array and Tournament Tree. What are minimum number of comparisons do we need (optimal number of calls to HaveAcquaintance())?

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