Maximum cells attacked by Rook or Bishop in given Chessboard

Given three integers N, R, and C representing an N*N chessboard and the position (R, C) where the rook and the bishop is placed. The task is to find out who can attack the most number of cells (except the cell they are in) and how many.

Note:

• A rook can move only horizontally along the row or vertically along the column and any number of cells at a time
• A bishop can move diagonally any number of cells at a time.

Examples:

Input: N = 3, R = 2, C = 1
Output: Rook, 4
Explanation: Rook can attack 2 cells in the row and 2 cells along the column. So 2+2 = 4.
Bishop can attack only 2 cells (1, 2) and (3, 2).

Input: N=1, R=1, C=1
Output: Both, 0

Approach: The problem can be solved by the following observation:

A rook can move vertically upwards or downwards and horizontally to the left or to the right. 
So total cells attacked by rook = (N – 1) + (N – 1) = 2*(N – 1)

A bishop can attack only diagonally i.e., across the primary diagonal or the secondary diagonal.
So total number of cells attacked along the main diagonal is min(R-1, C-1) + min(N-R, N-C).
Total number of cells attacked along the secondary diagonal is min(R-1, N-C) + min(N-R, C-1)

Follow the steps below to solve the problem:

• Find the total cells under attack by the rook (say X) using the above formula.
• Find the total cells under attack by the bishop (say Y) using the above formula.
• Check which one is greater and return the answer accordingly.

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach

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

// Function to find No of cells Elephant
// and Camel can attack and return max of that
pair<int, int> fun(int n, int r, int c)
{

    // Edge Case
    if (n == 1)
        return make_pair(0, 0);

    // For Rook
    int row = n - 1, col = n - 1;

    // For Bishop
    int UL = min(r - 1, c - 1);
    int UR = min(r - 1, n - c);
    int DL = min(n - r, c - 1);
    int DR = min(n - r, n - c);

    // Count total moves of Rook
    int E = row + col;

    // Count total moves of Bishop
    int C = DL + DR + UL + UR;

    // Return maximum among two, consider
    // 0 for Rook, 1 for Bishop, -1 for both
    if (E == C)
        return { -1, E };
    if (E > C)
        return { 0, E };
    return { 1, C };
}

// Driver Code
int main()
{
    int N = 3, R = 2, C = 1;

    // Function call
    pair<int, int> p = fun(N, R, C);

    if (p.first == -1)
        cout << "Both, ";
    else if (p.first == 0)
        cout << "Rook, ";
    else
        cout << "Bishop, ";
    cout << p.second << endl;
    return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
  // Function to find No of cells Elephant
  // and Camel can attack and return max of that
  static int[] fun(int n, int r, int c)
  {

    int[] res = new int[2];
    // Edge Case
    if (n == 1){
      res[0] = 0;
      res[1] = 0;
      return res;
    }

    // For Rook
    int row = n - 1, col = n - 1;

    // For Bishop
    int UL = Math.min(r - 1, c - 1);
    int UR = Math.min(r - 1, n - c);
    int DL = Math.min(n - r, c - 1);
    int DR = Math.min(n - r, n - c);

    // Count total moves of Rook
    int E = row + col;

    // Count total moves of Bishop
    int C = DL + DR + UL + UR;

    // Return maximum among two, consider
    // 0 for Rook, 1 for Bishop, -1 for both
    if (E == C){
      res[0] = -1;
      res[1] = E;
    }
    else if (E > C){
      res[0] = 0;
      res[1] = E;
    }
    else{
      res[0] = 1;
      res[1] = C;
    }
    return res;
  }

  // Driver code
  public static void main(String args[])
  {
    int N = 3, R = 2, C = 1;

    // Function call
    int[] p = fun(N, R, C);

    if (p[0] == -1)
      System.out.print("Both, ");
    else if (p[0] == 0)
      System.out.print("Rook, ");
    else
      System.out.print("Bishop, ");
    System.out.println(p[1]);
  }
}

// This code is contributed by shinjanpatra

Python3

# Python program for the above approach

# Function to find No of cells Elephant
# and Camel can attack and return max of that
def fun(n, r, c):

    # Edge Case
    if (n == 1):
        return [0, 0];

    # For Rook
    row = n - 1
    col = n - 1

    # For Bishop
    UL = min(r - 1, c - 1);
    UR = min(r - 1, n - c);
    DL = min(n - r, c - 1);
    DR = min(n - r, n - c);

    # Count total moves of Rook
    E = row + col;

    # Count total moves of Bishop
    C = DL + DR + UL + UR;

    # Return maximum among two, consider
    # 0 for Rook, 1 for Bishop, -1 for both
    if (E == C):
        return [-1, E];
    if (E > C):
        return [0, E];
    return [1, C];


# Driver Code

N = 3
R = 2
C = 1

# Function call
p = fun(N, R, C);

if (p[0] == -1):
    print("Both, ", end="");
elif (p[0] == 0):
    print("Rook, ", end="");
else:
    print("Bishop, ", end="");
print(p[1]);

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;

public class GFG
{

  // Function to find No of cells Elephant
  // and Camel can attack and return max of that
  static int[] fun(int n, int r, int c)
  {

    int[] res = new int[2];

    // Edge Case
    if (n == 1){
      res[0] = 0;
      res[1] = 0;
      return res;
    }

    // For Rook
    int row = n - 1, col = n - 1;

    // For Bishop
    int UL = Math.Min(r - 1, c - 1);
    int UR = Math.Min(r - 1, n - c);
    int DL = Math.Min(n - r, c - 1);
    int DR = Math.Min(n - r, n - c);

    // Count total moves of Rook
    int E = row + col;

    // Count total moves of Bishop
    int C = DL + DR + UL + UR;

    // Return maximum among two, consider
    // 0 for Rook, 1 for Bishop, -1 for both
    if (E == C){
      res[0] = -1;
      res[1] = E;
    }
    else if (E > C){
      res[0] = 0;
      res[1] = E;
    }
    else{
      res[0] = 1;
      res[1] = C;
    }
    return res;
  }

  // Driver code
  public static void Main(String []args)
  {
    int N = 3, R = 2, C = 1;

    // Function call
    int[] p = fun(N, R, C);

    if (p[0] == -1)
      Console.Write("Both, ");
    else if (p[0] == 0)
      Console.Write("Rook, ");
    else
      Console.Write("Bishop, ");

    Console.WriteLine(p[1]);
  }
}

// This code is contributed by AnkThon

Javascript

<script>
       // JavaScript program for the above approach

       // Function to find No of cells Elephant
       // and Camel can attack and return max of that
       function fun(n, r, c) {

           // Edge Case
           if (n == 1)
               return [0, 0];

           // For Rook
           let row = n - 1, col = n - 1;

           // For Bishop
           let UL = Math.min(r - 1, c - 1);
           let UR = Math.min(r - 1, n - c);
           let DL = Math.min(n - r, c - 1);
           let DR = Math.min(n - r, n - c);

           // Count total moves of Rook
           let E = row + col;

           // Count total moves of Bishop
           let C = DL + DR + UL + UR;

           // Return maximum among two, consider
           // 0 for Rook, 1 for Bishop, -1 for both
           if (E == C)
               return [-1, E];
           if (E > C)
               return [0, E];
           return [1, C];

       }
       // Driver Code

       let N = 3, R = 2, C = 1;

       // Function call
       let p = fun(N, R, C);

       if (p[0] == -1)
           document.write("Both, ");
       else if (p[0] == 0)
           document.write("Rook, ");
       else
           document.write("Bishop, ");
       document.write(p[1] + '<br>');

   // This code is contributed by Potta Lokesh

   </script>

Output

Rook, 4

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