Four Substrings to Matrix

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 Four Substrings to Matrix 

PROBLEM STATEMENT :

The program must accept N string values of equal length L as the input. The program must divide each string into four substrings of equal length. Then the program must form a character matrix of size R*C, where R = (N*3) and C = (L/4)*3. Then the program must fill the character matrix based on the following conditions. 

- The top-left, top-right, bottom-left, bottom-right and middle submatrices of size (R/3)*(C/3) must be filled with the asterisks. 

- The top-middle submatrix of size (R/3)*(C/3) must be filled with the 1st substrings of the N string values. 

- The middle-right submatrix of size (R/3)*(C/3) must be filled with the 2nd substrings of the N string values. 

- The bottom-middle submatrix of size (R/3)*(C/3) must be filled with the 3rd substrings of the N string values. 

- The middle-left submatrix of size (R/3)*(C/3) must be filled with the 4th substrings of the N string values. 

Finally, the program must print the R*C character matrix as the output. 

Note: The value of L is always divisible by 4. 

Boundary Condition(s): 

1 <= N <= 50 

4 <= L <= 100 

Input Format: 

The first line contains N. 

The next N lines, each contains a string. 

Output Format: 

The first R lines, each contains C characters representing the R*C character matrix. 

Example Input/Output 1: 
Input: () 

4 
abcdefghijklmnop 
ABCDEFGHIJKLMNOP 
1234567891011121 
zyxwvutsrqponmlk 
Output: 
****abcd**** 
****ABCD**** 
****1234**** 
****zyxw**** 
mnop****efgh 
MNOP****EFGH 
1121****5678 
nmlk****vuts 
****ijkl**** 
****IJKL**** 
****9101**** 
****rqpo****

Explanation: 

Here N = 4 and L = 16. 

The size of the character matrix to be formed is 12*12 (where R = 4*3 and C = (16/4)*3). 

Then the matrix is filled with the characters based on the given conditions. 

Example Input/Output 2: 
Input: () 

7 
interest 
absolute 
complete 
accurate 
delivery 
achieved 
dominant 
Output: 
**in** 
**ab** 
**co** 
**ac** 
**de** 
**ac** 
**do** 
st**te 
te**so 
te**mp 
te**cu 
ry**li 
ed**hi 
nt**mi 
**re** 
**lu** 
**le** 
**ra** 
**ve** 
**ev** 
**na**


SOLUTION :

C (Programming Language)


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

int main()
{
    int n,i,j,m;
    scanf("%d\n",&n);
    char s[n][101];
    for(i=0;i<n;i++)
    scanf("%s\n",s[i]);
    m=strlen(s[0])/4;
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
        printf("*");
        for(j=0;j<m;j++)
        printf("%c",s[i][j]);
        for(j=0;j<m;j++)
        printf("*");
        printf("\n");
    }
    for(i=0;i<n;i++)
    {
        for(j=m*3;j<m*4;j++)
        printf("%c",s[i][j]);
        for(j=0;j<m;j++)
        printf("*");
        for(j=m;j<m*2;j++)
        printf("%c",s[i][j]);
        printf("\n");
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<m;j++)
        printf("*");
        for(j=m*2;j<m*3;j++)
        printf("%c",s[i][j]);
        for(j=0;j<m;j++)
        printf("*");
        printf("\n");
    }

}


C++ (CPP)

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

int main(int argc, char** argv)
{
    
    int n ;
    
    cin >> n;
    
    string arr[n];
    
    for(int i = 0 ; i<n ; i++)
       cin >> arr[i];
       
    int len = arr[0].size() , r = n*3 , c = (len/4)*3 , t = len/4;
    
    string star(t , '*');
    
    for(int j = 0 ; j<n ; j++)
    {
        cout << star;
        cout << arr[j].substr(0,t);
        arr[j] = arr[j].substr(t);
        cout << star << endl;
    }
    
    for(int i = 0 ; i<n ; i++)
    {
        cout << arr[i].substr(arr[i].size()-t);
        arr[i] = arr[i].substr(0,arr[i].size()-t);
        cout << star;
        cout << arr[i].substr(0,t) << endl;
        arr[i] = arr[i].substr(t);
    }
    
    for(int i = 0 ; i<n ; i++)
    {
        cout << star;
        cout << arr[i].substr(0,t);
        cout << star << endl;
    }

}


JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int N=sc.nextInt();sc.nextLine();
		String s=sc.nextLine();
		int len=s.length()/4;
		String[][] str=new String[N][4];
		str[0][0]=s.substring(0,len);
		str[0][1]=s.substring(len,len*2);
		str[0][2]=s.substring(len*2,len*3);
		str[0][3]=s.substring(len*3,len*4);
		for(int i=1;i<N;i++)
		{
		    String t=sc.nextLine();
		    str[i][0]=t.substring(0,len);
		    str[i][1]=t.substring(len,len*2);
		    str[i][2]=t.substring(len*2,len*3);
		    str[i][3]=t.substring(len*3,len*4);
		}
		for(int i=0;i<N;i++)
		{
		    for(int j=0;j<len;j++)
		    System.out.print("*");
		    System.out.print(str[i][0]);
		    for(int j=0;j<len;j++)
		    System.out.print("*");
		    System.out.println();
		}
		for(int i=0;i<N;i++)
		{
		    System.out.print(str[i][3]);
		    for(int j=0;j<len;j++)
		    System.out.print("*");
		    System.out.print(str[i][1]);
		    System.out.println();
		}
		for(int i=0;i<N;i++)
		{
		    for(int j=0;j<len;j++)
		    System.out.print("*");
		    System.out.print(str[i][2]);
		    for(int j=0;j<len;j++)
		    System.out.print("*");
		    System.out.println();
		}
	}
}


PYTHON

       
n=int(input())
l=[input().strip() for i in range(n)]
k=len(l[0])//4
for i in range(n):
    l[i]=[l[i][:k],l[i][k:2*k],l[i][2*k:3*k],l[i][3*k:4*k]]
res=[]
for i in range(n):
    res.append(('*'*k)+l[i][0]+('*'*k))
for i in range(n):
    res.append(l[i][3]+('*'*k)+l[i][1])
for i in range(n):
    res.append(('*'*k)+l[i][2]+('*'*k))
for i in res:
    print(i)


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