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The Celebrity Problem | Set-2

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  • Last Updated : 11 Jan, 2022

In a party of N people, only one person is known to everyone. Such a person may be present at 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 the problem be solved? 

Also given a list know[], where know[i] is represented in form {a, b} means person a knows person b.

Examples: 

Input: know[] = {{1, 3}, {2, 3}, {4, 3}}, N = 4
Output: 2
Explanation: Person with id = 3 does not know anyone, but every other person knows him.

Input: know[] = {{1, 3}, {2, 3}, {3, 2}, {4, 3}}, N = 4
Output: -1
Explanation: No celebrity is present in the party.

 

Approach: The naive approach and some optimized approaches are discussed in Set-1 of this problem.

Greedy Approach: It can be solved using greedy approach. The above problem can be solved by creating a vector of bool, and traversing the given matrix as follows:

  • Traverse the matrix once and find the person that knows no one.
  • Store this as the current celebrity. If current celebrity is no one, then return -1, else go to the next step.
  • Traverse the matrix again and now check if every person at the party knows the current celebrity.
  • If the current celebrity meets this criterion as well, return it as the party celebrity.
  • Else return -1.

Below is the implementation of the above approach

C++




// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the celebrity
int findPartyCelebrity(int N,
                       vector<vector<int> >& know)
{
 
    vector<bool> celebrity(N + 1, false);
    for (int i = 0; i < know.size(); i++)
        celebrity[know[i][0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
        if (celebrity[i] == false)
            party_celebrity = i;
 
    celebrity.assign(N + 1, false);
    for (int i = 0; i < know.size(); i++)
        if (know[i][1] == party_celebrity)
            celebrity[know[i][0]] = true;
 
    for (int i = 1; i <= N; i++)
        if (i != party_celebrity && celebrity[i] == false)
            party_celebrity = -1;
 
    return party_celebrity;
}
 
// Driver code
int main()
{
    int N = 4;
    vector<vector<int> > know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
    cout << findPartyCelebrity(N, know);
    return 0;
}


Java




// Java code for the above approach
import java.io.*;
 
class GFG {
 
  // Function to find the celebrity
  static int findPartyCelebrity(int N, int[][] know)
  {
 
    boolean[] celebrity = new boolean[N + 1];
    for (int i = 0; i < know.length; i++)
      celebrity[know[i][0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
      if (celebrity[i] == false)
        party_celebrity = i;
 
    celebrity = new boolean[N + 1];
    for (int i = 0; i < know.length; i++)
      if (know[i][1] == party_celebrity)
        celebrity[know[i][0]] = true;
 
    for (int i = 1; i <= N; i++)
      if (i != party_celebrity
          && celebrity[i] == false)
        party_celebrity = -1;
 
    return party_celebrity;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 4;
    int[][] know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
    System.out.println(findPartyCelebrity(N, know));
  }
}
 
// This code is contributed by Potta Lokesh


Python3




# Python code to implement above approach
 
# Function to find the celebrity
def findPartyCelebrity(N, know):
 
    celebrity = [False] * (N + 1)
    for i in range(len(know)):
        celebrity[know[i][0]] = True
 
    party_celebrity = -1
    for i in range(1, N + 1):
        if celebrity[i] == False:
            party_celebrity = i
 
    celebrity = [False] * (N + 1)
    for i in range(len(know)):
        if know[i][1] == party_celebrity:
            celebrity[know[i][0]] = True
 
    for i in range(1, N + 1):
        if i != party_celebrity and celebrity[i] == False:
            party_celebrity = -1
 
    return party_celebrity
 
# Driver code
N = 4
know = [[1, 3], [2, 3], [4, 3]]
print(findPartyCelebrity(N, know))
 
# This code is contributed by gfgking


C#




// C# code for the above approach
using System;
 
class GFG
{
 
  // Function to find the celebrity
  static int findPartyCelebrity(int N, int[,] know)
  {
 
    bool[] celebrity = new bool[N + 1];
    for (int i = 0; i < know.GetLength(0); i++)
      celebrity[know[i, 0]] = true;
 
    int party_celebrity = -1;
    for (int i = 1; i <= N; i++)
      if (celebrity[i] == false)
        party_celebrity = i;
     
    for(int i = 0; i < N + 1; i++){
        celebrity[i] = false;
    }
     
    for (int i = 0; i < know.GetLength(0); i++)
      if (know[i, 1] == party_celebrity)
        celebrity[know[i, 0]] = true;
 
    for (int i = 1; i <= N; i++)
      if (i != party_celebrity
          && celebrity[i] == false)
        party_celebrity = -1;
 
    return party_celebrity;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 4;
    int[,] know = { { 1, 3 }, { 2, 3 }, { 4, 3 } };
        Console.Write(findPartyCelebrity(N, know));
  }
}
 
// This code is contributed by Samim Hossain Mondal


Javascript




<script>
    // JavaScript code to implement above approach
 
    // Function to find the celebrity
    const findPartyCelebrity = (N, know) => {
 
        let celebrity = new Array(N + 1).fill(false);
        for (let i = 0; i < know.length; i++)
            celebrity[know[i][0]] = true;
 
        let party_celebrity = -1;
        for (let i = 1; i <= N; i++)
            if (celebrity[i] == false)
                party_celebrity = i;
 
        celebrity = new Array(N + 1).fill(false);
        for (let i = 0; i < know.length; i++)
            if (know[i][1] == party_celebrity)
                celebrity[know[i][0]] = true;
 
        for (let i = 1; i <= N; i++)
            if (i != party_celebrity && celebrity[i] == false)
                party_celebrity = -1;
 
        return party_celebrity;
    }
 
    // Driver code
    let N = 4;
    let know = [[1, 3], [2, 3], [4, 3]];
    document.write(findPartyCelebrity(N, know));
 
// This code is contributed by rakeshsahni
 
</script>


 
 

Output: 

3

 

Time Complexity: O(N)
Auxiliary Space: O(N) 


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