Knight - Possible Moves

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 Knight - Possible Moves

PROBLEM STATEMENT :

The program must accept the position of a Knight (Horse) in a chessboard denoted by the letter 'H' and print the number of moves M it can make from that position. Assume no other pieces are on the chessboard which is denoted
by the letter B.

Note: In Chess, a knight can move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an L).

Input Format:
8 lines and one line among them containing the letter H.

Output Format:
The first line contains M.


Example Input/Output 1: 
Input: () 

B B B B B B H B
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B 
Output: 
3

Explanation: 

The positions where the Knight can move are indicated by '*'.

B B B B B B H B
B B B B * B B B
B B B B B * B *
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B


As three moves are possible, 3 is printed as the output.

Example Input/Output 2: 
Input: () 

B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B H B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B
B B B B B B B B 
Output: 
8



       
                    


            1)    LEARN THRICE 

                                👇 

            2)    THINK TWICE

                                👇 

            3)    APPLY ONCE




SOLUTION :

C (Programming Language)


      
#include<stdio.h>
#include<stdlib.h>

int main()
{
    char arr[8][8];
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<8;j++)
        {
            scanf("%c ",&arr[i][j]);
        }
    }
    int count=0;
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<8;j++)
        {
            if(arr[i][j]=='H')
            {
                int moves[8][2]={{-1,2},{1,2},{-1,-2},{1,-2},{2,1},{2,-1},{-2,1},{-2,-1}};
                for(int k=0;k<8;k++)
                {
                    if((i+moves[k][0])>=0 && (i+moves[k][0])<8 && (j+moves[k][1])>=0 && (j+moves[k][1])<8)
                    {
                        count++;
                    }
                }
                printf("%d",count);
                return 0;
            }
        }
    }
}


C++ (CPP)

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

int main(int argc, char** argv)
{
    char arr[8][8];
    int x, y;
    for(int i=0; i<8; i++){
        for(int j=0; j<8; j++){
            cin >> arr[i][j];
            if(arr[i][j]=='H'){
                x=i;
                y=j;
            }
        }
    }
    int dir[8][2]={{-1,-2},{-2,-1},{-1,2},{-2,1},{1,-2},{2,-1},{1,2},{2,1}};
    int ans=0;
    for(int i=0; i<8; i++){
        int sr=x+dir[i][0], sc=y+dir[i][1];
        if(sr<0 || sc<0 || sr>=8 || sc>=8) continue;
        ans++;
    }
    cout << ans;
}


JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
	Scanner in=new Scanner(System.in);
	char mat[][]=new char[8][8];
	int kx=0;
	int ky=0;
	for(int i=0;i<8;i++)
	{
    		for(int j=0;j<8;j++)
    		{
        		char ch=in.next().charAt(0);
        		mat[i][j]=ch;
        		if(ch=='H')
        		{
            		kx=i;ky=j;
        		}
    		}
	}
	int co=0;
	int dir1[]={-2,-2,2,2,-1,-1,1,1};
	int dir2[]={-1,1,-1,1,-2,2,-2,2};
	for(int i=0;i<dir1.length;i++)
	{
    		int nx=dir1[i]+kx;
    		int ny=dir2[i]+ky;
    		if(ny<0 || nx<0 || nx>=8 || ny>=8)
    			continue;
    		co++;
	}
	System.out.println(co);
   }
}


PYTHON

       
L=[input().strip().split() for i in range(8)]
X=[[2,-1],[2,1],[-2,1],[-2,-1],[-1,2],[1,2],[-1,-2],[1,-2]]
for i in range(8):
    for j in range(8):
        if L[i][j]=='H':
            TotalCount=0
            for x,y in X:
                if 0<=i+x<8 and 0<=j+y<8: TotalCount+=1
print(TotalCount)


Never Stop Learning !!


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