Construction of New Buildings

Construction of New Buildings 

PROBLEM STATEMENT :

The program must accept a character of size R*C representing a city as the input. The matrix consists of asterisks and hyphens. Each asterisk represents a building and each hyphen represents a land. The government has planned to construct new buildings in the city based on the following condition. 

- If a land is present in one of the four corners (top-left, top-right, bottom-left and bottom-right) of a building, then a new building is allowed to construct on that land. 

The program must print the maximum number of new buildings that can be built based on the given conditions. 

Boundary Condition(s): 

2 <= R, C <= 50 

Input Format: 

The first line contains R and C separated by a space. 

The next R lines, each contains C characters separated by a space. 

Output Format: 

The first line contains the maximum number of new buildings that can be built. 

Example Input/Output 1: 
Input: () 
5 6 
- - * - * - 
- * - - - * 
* - - - - * 
- * - - - * 
* * * * - -  
Output: 
10 

Explanation: 

The 10 new buildings can be constructed as given below. 

# - * - * - 

- * - # # * 

* - # - # * 

# * # # # * 

* * * * # - 

Each hash symbol in the above matrix represents a new building. 

So 10 is printed as the output.  

Example Input/Output 2: 
Input: () 
3 3 
- * - 
* - * 
- * - 
Output: 
0 

---

SOLUTION :

C (Programming Language)

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

int main()
{
    int cnt=0,r,c,x,i,j,i1,j1,a1[]={1,1,-1,-1},a2[]={1,-1,1,-1};
    scanf("%d%d",&r,&c);
    char ch[r][c],s[2];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            scanf("%s",s);
            ch[i][j]=s[0];
        }
    }
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            if(ch[i][j]=='*')
            {
                for(x=0;x<4;++x)
                {
                    i1=i+a1[x];
                    j1=j+a2[x];
                    if((i1>=0&&i1<r&&j1>=0&&j1<c)&&ch[i1][j1]=='-')
                    {
                        ch[i1][j1]='#';
                        cnt++;
                    }
                }
            }
        }
    }
    printf("%d",cnt);
}
 

C++ (CPP)

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

int main(int argc, char** argv)
{
    int r,c,cnt=0,i,j,x,i1,j1,a1[]={1,1,-1,-1},a2[]={1,-1,1,-1};
    cin>>r>>c;
    char ch[r][c];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            cin>>ch[i][j];
        }
    }
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            if(ch[i][j]=='*')
            {
                for(x=0;x<4;++x)
                {
                    i1=i+a1[x];
                    j1=j+a2[x];
                    if((i1>=0&&i1<r&&j1>=0&&j1<c)&&ch[i1][j1]=='-')
                    {
                        ch[i1][j1]='#';
                        cnt++;
                    }
                }
            }
        }
    }
    cout<<cnt;
}
 

JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int r=sc.nextInt(),c=sc.nextInt(),i,j,i1,j1,x,cnt=0;
		int a1[]={1,1,-1,-1}, a2[]={1,-1,1,-1};
		char[][] ch=new char[r][c];
		for(i=0;i<r;++i)
		{
		    for(j=0;j<c;++j)
		    {
		        ch[i][j]=sc.next().charAt(0);
		    }
		}
		for(i=0;i<r;++i)
		{
		    for(j=0;j<c;++j)
		    {
		        if(ch[i][j]=='*')
		        {
		            for(x=0;x<4;++x)
		            {
		                i1=i+a1[x];
		                j1=j+a2[x];
		                if((i1>=0&&i1<r&&j1>=0&&j1<c)&&ch[i1][j1]=='-')
		                {
		                    ch[i1][j1]='#';
		                    cnt++;
		                }
		            }
		        }
		    }
		}
		System.out.println(cnt);
	}
}
 

PYTHON

       
r,c=map(int,input().split())
cnt=0
l=[input().strip().split() for i in range(r)]
for i in range(r):
    for j in range(c):
        dir1=[-1,1,-1,1]
        dir2=[-1,1,1,-1]
        if l[i][j]=='*':
            for x in range(4):
                i1=i+dir1[x]
                j1=j+dir2[x]
                if((i1>=0 and i1<r and j1>=0 and j1<c) and l[i1][j1]=='-'):
                    l[i1][j1]='#'
                    cnt+=1
print(cnt)
 

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