Moving on grid

Given a grid on the XY plane with dimensions r x c (where r denotes maximum cells along the X axis and c denotes maximum cells along the Y axis), the two players (say JON and ARYA ) can move the coin over the grid satisfying the following rules:

• There is a coin on (1, 1) cell initially.
• JON will move first.
• Both will play on alternate moves.
• In each move, they can place the coin in the following positions if the current position of the coin is x, y
• (x+1, y), (x+2, y), (x+3, y), (x, y+1), (x, y+2), (x, y+3), (x, y+4), (x, y+5), (x, y+6)
• They can't go outside the grid.
• Player who cannot make any move will lose this game.
• Both play optimally.

Examples:

Input: r = 1, c = 2
Output: JON 
Explanation: ARYA lost the game because
he won't able to move after JON's move. 

Input: r = 2, c = 2
Output: ARYA
Explanation: After first move by JON (1, 2 or 2, 1)
and second move by ARYA(2, 2) JON won't able to
move so ARYA wins. 

Approach: This problem can be solved using game theory based on the following idea:

Check the following points:

• For rows whoever leaves 4 cells to be covered wins the game and for columns, whoever leaves 7 cells to be covered, wins the game.
• To win the game one has to make sure that the opponent cannot win either row or column i.e. he can win both the row and column and leaves 4 cells in row and 7 cells in columns.
• The game starts with Jon. So if (r-1)%7 = (c-1)%4, then Jon can win either row or column but not both. So Arya wins that game.
• In all other cases Jon wins the game.

Follow the steps to solve the problem:

• Get the value of (r-1)%7 and (c-1)%4.
• If these two values are the same, then Arya wins. 
• Otherwise, Jon wins.

Below is the implementation of the above approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the winner
string moveOnGrid(int r, int c)
{
    r = (r - 1) % 7;
    c = (c - 1) % 4;
    if (r != c)
        return "JON";
    return "ARYA";
}

// Driver code
int main()
{
    int r = 2, c = 2;

    // Function call
    cout << moveOnGrid(r, c);
    return 0;
}

// Java code to implement the approach
import java.io.*;

class GFG
{

  // Function to find the winner
  public static String moveOnGrid(int r, int c)
  {
    r = (r - 1) % 7;
    c = (c - 1) % 4;
    if (r != c)
      return "JON";
    return "ARYA";
  }

  // Driver Code
  public static void main(String[] args)
  {
    int r = 2, c = 2;

    // Function call
    System.out.print(moveOnGrid(r, c));
  }
}

// This code is contributed by Rohit Pradhan

# Python code to implement the approach

# Function to find the winner
def moveOnGrid(r, c):

    r = (r - 1) % 7
    c = (c - 1) % 4
    if (r != c):
        return "JON"
    return "ARYA"

# Driver code
r,c = 2,2

# Function call
print(moveOnGrid(r, c))

# This code is contributed by shinjanpatra

// C# code to implement the approach
using System;
class GFG {

  // Function to find the winner
  public static String moveOnGrid(int r, int c)
  {
    r = (r - 1) % 7;
    c = (c - 1) % 4;
    if (r != c)
      return "JON";
    return "ARYA";
  }

  // Driver Code
  public static void Main ()
  {
    int r = 2, c = 2;

    // Function call
    Console.Write(moveOnGrid(r, c));
  }
}

// This code is contributed by jana_sayantan.

<script>

// JavaScript code to implement the approach

// Function to find the winner
function moveOnGrid(r, c)
{
    r = (r - 1) % 7;
    c = (c - 1) % 4;
    if (r != c)
        return "JON";
    return "ARYA";
}

// Driver code
let r = 2, c = 2;

// Function call
document.write(moveOnGrid(r, c));

// This code is contributed by shinjanpatra

</script>

ARYA

Time Complexity: O(1)
Auxiliary Space: O(1)