// 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