Count the total number of squares that can be visited by Bishop in one move

Given the position of a Bishop on an 8 * 8 chessboard, the task is to count the total number of squares that can be visited by the Bishop in one move. The position of the Bishop is denoted using row and column number of the chessboard. 

Examples:

Input: Row = 4, Column = 4 
Output: 13

Input: Row = 1, Column = 1 
Output: 7 

Approach: In the game of chess, a Bishop can only move diagonally and there is no restriction in distance for each move.

So, We can also say that Bishop can move in four ways i.e. diagonally top left, top right, bottom left and bottom right from current position.

We can calculate the numbers of squares visited in each move by: 

Total squares visited in Top Left move = min(r, c) – 1 
Total squares visited in Top Right move = min(r, 9 – c) – 1 
Total squares visited in Bottom Left move = 8 – max(r, 9 – c) 
Total squares visited in Bottom Right move = 8 – max(r, c) 
where, r and c are the coordinates of the current position of the Bishop on the chessboard. 

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of
// total positions the Bishop
// can visit in a single move
int countSquares(int row, int column)
{

    // Count top left squares
    int topLeft = min(row, column) - 1;

    // Count bottom right squares
    int bottomRight = 8 - max(row, column);

    // Count top right squares
    int topRight = min(row, 9 - column) - 1;

    // Count bottom left squares
    int bottomLeft = 8 - max(row, 9 - column);

    // Return total count
    return (topLeft + topRight + bottomRight + bottomLeft);
}

// Driver code
int main()
{

    // Bishop's Position
    int row = 4, column = 4;

    cout << countSquares(row, column);

    return 0;
}

Java

// Java implementation of above approach
class GFG {

    // Function to return the count of
    // total positions the Bishop
    // can visit in a single move
    static int countSquares(int row, int column)
    {

        // Count top left squares
        int topLeft = Math.min(row, column) - 1;

        // Count bottom right squares
        int bottomRight = 8 - Math.max(row, column);

        // Count top right squares
        int topRight = Math.min(row, 9 - column) - 1;

        // Count bottom left squares
        int bottomLeft = 8 - Math.max(row, 9 - column);

        // Return total count
        return (topLeft + topRight + bottomRight + bottomLeft);
    }

    // Driver code
    public static void main(String[] args)
    {

        // Bishop's Position
        int row = 4, column = 4;

        System.out.println(countSquares(row, column));
    }
}

C#

// C# implementation of above approach
using System;
class GFG {

    // Function to return the count of
    // total positions the Bishop
    // can visit in a single move
    static int countSquares(int row, int column)
    {

        // Count top left squares
        int topLeft = Math.Min(row, column) - 1;

        // Count bottom right squares
        int bottomRight = 8 - Math.Max(row, column);

        // Count top right squares
        int topRight = Math.Min(row, 9 - column) - 1;

        // Count bottom left squares
        int bottomLeft = 8 - Math.Max(row, 9 - column);

        // Return total count
        return (topLeft + topRight + bottomRight + bottomLeft);
    }

    // Driver code
    public static void Main()
    {

        // Bishop's Position
        int row = 4, column = 4;

        Console.WriteLine(countSquares(row, column));
    }
}

Python3

# Python3 implementation of above approach

# Function to return the count of
# total positions the Bishop
# can visit in a single move
def countSquares(row, column):

    # Count top left squares
    topLeft = min(row, column) - 1

    # Count bottom right squares
    bottomRight = 8 - max(row, column)

    # Count top right squares
    topRight = min(row, 9-column) -1

    # Count bottom left squares
    bottomLeft = 8 - max(row, 9-column)


    # Return total count
    return (topLeft + topRight + bottomRight + bottomLeft)

# Driver code

# Bishop's Position
row = 4
column = 4

print(countSquares(row, column))

PHP

<?php
// PHP implementation of above approach

// Function to return the count of
// total positions the Bishop
// can visit in a single move
function countSquares($row, $column)
{

    // Count top left squares
    $topLeft = min($row, $column) - 1;

    // Count bottom right squares
    $bottomRight = 8 - max($row, $column);

    // Count top right squares
    $topRight = min($row, 9 - $column) - 1;

    // Count bottom left squares
    $bottomLeft = 8 - max($row, 9 - $column);

    // Return total count
    return ($topLeft + $topRight +
            $bottomRight + $bottomLeft);
}

// Driver code

// Bishop's Position
$row = 4;
$column = 4;
echo countSquares($row, $column);

// This code is contributed by jit_t
?>

Javascript

<script>

// Javascript implementation of above approach

// Function to return the count of
// total positions the Bishop
// can visit in a single move
function countSquares(row, column)
{

    // Count top left squares
    var topLeft = Math.min(row, column) - 1;

    // Count bottom right squares
    var bottomRight = 8 - Math.max(row, column);

    // Count top right squares
    var topRight = Math.min(row, 9 - column) - 1;

    // Count bottom left squares
    var bottomLeft = 8 - Math.max(row, 9 - column);

    // Return total count
    return (topLeft + topRight + bottomRight + bottomLeft);
}

// Driver code
// Bishop's Position
var row = 4, column = 4;
document.write( countSquares(row, column));

</script>

Output

13

Complexity Analysis:

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