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Total position where king can reach on a chessboard in exactly M moves | Set 2

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  • Last Updated : 09 Jun, 2022

Given the position of the king on an 8 X 8 chessboard, the task is to count the total number of squares that can be visited by the king in m moves. The position of the king is denoted using row and column number. 
Note: The square which is currently acquired by the king is already visited and will be counted in the result.
Examples: 
 

Input: r = 4, c = 4, m = 1 
Output: 9
Input: r = 4, c = 4, m = 2 
Output: 25 

Approach: A king can move one square in any direction (i.e horizontally, vertically and diagonally). So, in one move king can visit its adjacent squares. 
 

King on chess

So, A square which is within m units distance (Considering 1 Square as 1 unit distance) from the king’s current position can be visited in m moves. 
For all squares of the chessboard, check if a particular square is at m unit distance away or less from King’s current position.

  1. Increment count, if step 1 is true.
  2. Print the count

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of squares
// that can be visited by king in m moves
int countSquares(int r, int c, int m)
{
 
    // To store the count of squares
    int squares = 0;
 
    // Check all squares of
    // the chessboard
    for (int i = 1; i <= 8; i++) {
        for (int j = 1; j <= 8; j++) {
 
            // Check if square (i, j) is
            // at a distance <= m units
            // from king's current position
            if (max(abs(i - r), abs(j - c)) <= m)
                squares++;
        }
    }
 
    // Return count of squares
    return squares;
}
 
// Driver code
int main()
{
    int r = 4, c = 4, m = 1;
 
    cout << countSquares(r, c, m) << endl;
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return the count of squares
    // that can be visited by king in m moves
    static int countSquares(int r, int c, int m)
    {
        // To store the count of squares
        int squares = 0;
 
        // Check all squares of
        // the chessboard
        for (int i = 1; i <= 8; i++) {
            for (int j = 1; j <= 8; j++) {
 
                // Check if square (i, j) is
                // at a distance <= m units
                // from king's current position
                if (Math.max(Math.abs(i - r), Math.abs(j - c)) <= m)
                    squares++;
            }
        }
 
        // Return count of squares
        return squares;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int r = 4, c = 4, m = 1;
        System.out.print(countSquares(r, c, m));
    }
}

C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the count of squares
    // that can be visited by king in m moves
    static int countSquares(int r, int c, int m)
    {
        // To store the count of squares
        int squares = 0;
 
        // Check all squares of
        // the chessboard
        for (int i = 1; i <= 8; i++) {
            for (int j = 1; j <= 8; j++) {
 
                // Check if square (i, j) is
                // at a distance <= m units
                // from king's current position
                if (Math.Max(Math.Abs(i - r), Math.Abs(j - c)) <= m)
                    squares++;
            }
        }
 
        // Return count of squares
        return squares;
    }
 
    // Driver code
    public static void Main()
    {
        int r = 4, c = 4, m = 1;
        Console.Write(countSquares(r, c, m));
    }
}

Python3




# Python implementation of the approach
 
# Function to return the count of squares
# that can be visited by king in m moves
def countSquares(r, c, m):
 
    # To store the count of squares
    squares = 0
     
    # Check all squares of
    # the chessboard
    for i in range (1, 9):
        for j in range (1, 9):
             
            # Check if square (i, j) is
            # at a distance <= m units
            # from king's current position
            if(max(abs(i - r), abs(j - c)) <= m):
                squares = squares + 1
         
    # Return count of squares
    return squares
 
# Driver code
r = 4
c = 4
m = 1
 
print(countSquares(r, c, m));

PHP




<?php
// PHP implementation of the approach
 
// Function to return the count of squares
// that can be visited by king in m moves
function countSquares($r, $c, $m)
{
 
    // To store the count of squares
    $squares = 0;
 
    // Check all squares of
    // the chessboard
    for ($i = 1; $i <= 8; $i++)
    {
        for ($j = 1; $j <= 8; $j++)
        {
 
            // Check if square (i, j) is
            // at a distance <= m units
            // from king's current position
            if (max(abs($i - $r),
                    abs($j - $c)) <= $m)
                $squares++;
        }
    }
 
    // Return count of squares
    return $squares;
}
 
// Driver code
$r = 4;
$c = 4;
$m = 1;
 
echo countSquares($r, $c, $m);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// Javascript implementation of the approach 
 
    // Function to return the count of squares
    // that can be visited by king in m moves
    function countSquares(r, c, m)
    {
        // To store the count of squares
        let squares = 0;
   
        // Check all squares of
        // the chessboard
        for (let i = 1; i <= 8; i++) {
            for (let j = 1; j <= 8; j++) {
   
                // Check if square (i, j) is
                // at a distance <= m units
                // from king's current position
                if (Math.max(Math.abs(i - r), Math.abs(j - c)) <= m)
                    squares++;
            }
        }
   
        // Return count of squares
        return squares;
    }
 
// Driver Code
 
         let r = 4, c = 4, m = 1;
        document.write(countSquares(r, c, m));
            
</script>

Output: 

9

 

Time Complexity: O(1), since the loop runs for a total of 64 times, that is constant time only.
Auxiliary Space: O(1), since no extra space has been taken.


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