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: 13Input: 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++ 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 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# 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 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 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 ?> |
<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> |
13
Complexity Analysis:
- Time Complexity: O(1)
- Auxiliary Space: O(1)