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Multi Layered Diamond Shaped Rhombus Pattern

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Multi Layered Diamond Shaped Rhombus Pattern

Given an odd value of N, the program must print multi layered rhombus pattern in diamond shapes whose side contains N, N-2, ... 1 slashes respectively as shown below in the examples.
Input Format:
The first line contains N.
Output Format:
The  multi layered rhombus pattern in diamond shapes whose side contains N, N-2, ... 1 slashes respectively. Hash symbol is used as a filler for other values.
Boundary Conditions:
1 <= N <= 101 and N is odd.
Example Input/Output 1:
Input:
5
Output:
####/\####
###/##\###
##/#/\#\##
#/#/##\#\#
/#/#/\#\#\
\#\#\/#/#/
#\#\##/#/#
##\#\/#/##
###\##/###
####\/####
Example Input/Output 2:
Input:
11
Output:
##########/\##########
#########/##\#########
########/#/\#\########
#######/#/##\#\#######
######/#/#/\#\#\######
#####/#/#/##\#\#\#####
####/#/#/#/\#\#\#\####
###/#/#/#/##\#\#\#\###
##/#/#/#/#/\#\#\#\#\##
#/#/#/#/#/##\#\#\#\#\#
/#/#/#/#/#/\#\#\#\#\#\
\#\#\#\#\#\/#/#/#/#/#/
#\#\#\#\#\##/#/#/#/#/#
##\#\#\#\#\/#/#/#/#/##
###\#\#\#\##/#/#/#/###
####\#\#\#\/#/#/#/####
#####\#\#\##/#/#/#####
######\#\#\/#/#/######
#######\#\##/#/#######
########\#\/#/########
#########\##/#########
##########\/##########

Code:
#include <iostream>
using namespace std;
int main(int argc, char** argv)
{
    int n,i,j;
    cin>>n;
    int x,y,f=1,f1=1,f2=1;
    x=n-1;
    y=n;
    char  arr[n*2][n*2];
    for(i=0;i<n;i++)
    {
        for(j=0;j<n*2;j++)
        {
            if(x==j&&f==1)
                {
                    //cout<<"/";
                    arr[i][j]='/';
                    x--;
                    f=0;
                }
            else if(y==j&&f1==1)
                {
                    arr[i][j]='\\';
                    //cout<<"\\";
                        y++;
                        f1=0;
                }
                else
                    arr[i][j]='#';
                    //cout<<"#";
            if(f2==0&&f1!=0)
            if(i>1)
            {
                arr[i][j]=arr[i-2][j];
            }
            if(f==0)
                f2=0;
        }

        f=1;f1=1;f2=1;
      
          
      
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n*2;j++)
        {
            cout<<arr[i][j];
    }
    cout<<"\n";
}
for(i=n-1;i>=0;i--)
    {
        for(j=0;j<n*2;j++)
        {
            if(arr[i][j]=='#')
            cout<<"#";
            else if(arr[i][j]=='/')
                cout<<"\\";
            else
                cout<<"/";
        }
        cout<<"\n";
    }


}

 Execute code Here:
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4 comments:

  1. Anonymous30 July 2017 at 09:59

    #include

    using namespace std;

    int main(int argc, char** argv)
    {
    char c[1000][1000],q,p;
    long long int n,m,i,j,k;
    cin>>n;
    for(i=0;i<2*n;i++)
    {
    for(j=0;j<2*n;j++)
    c[i][j]='#';
    }
    for(i=0;i<=n;i+=2)
    {
    c[i][n-1]='/';
    c[i][n]=92;
    }
    for(i=0;in;i++)
    {
    for(j=0;j<2*n;j++)
    {
    if(c[i][j]=92)
    {
    c[i-1][j-1]=92;
    }
    if(c[i][j]='/')
    c[i-1][j+1]='/';
    }
    }

    for(i=0;i<n*2;i++)
    {
    for(j=0;j<2*n;j++)
    cout<<c[i][j];

    cout<<endl;
    }
    }

    ReplyDelete
  2. Anonymous1 August 2017 at 18:39

    import java.util.*;
    public class Hello {

    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    String[][] matrix=new String[204][204];
    for(int i=0;i<2*n;i++)
    {
    for(int j=0;j<2*n;j++)
    {
    matrix[i][j]="#";
    }
    }

    for(int i=0;in;i--)
    {
    for(int j=0;j<2*n;j++)
    {
    if(matrix[i][j]=="\\")
    {
    matrix[i-1][j-1]="\\";

    }

    if(matrix[i][j]=="/")
    {
    matrix[i-1][j+1]="/";

    }
    }
    }


    for(int i=0;i<2*n;i++)
    {
    for(int j=0;j<2*n;j++)
    {
    System.out.print(matrix[i][j]);
    }
    System.out.println();
    }

    }
    }

    ReplyDelete
  3. Unknown2 August 2017 at 05:15

    Proper Java Implementation Of This Problem:
    Credits:Mr.Anonymous9896


    import java.util.*;
    public class Hello {

    public static void main(String[] args) {
    Scanner s= new Scanner(System.in);
    int num=s.nextInt();
    int x=num-1;
    int y=num;
    int f=1,f1=1,f2=1;
    String[][]str=new String[num*2][num*2];
    for(int i=0;i1)
    str[i][j]=str[i-2][j];
    }
    if(f==0)
    {
    f2=0;
    }
    }
    f=1;f1=1;f2=1;
    }
    for(int i=0;i=0;i--)
    {
    for(int j=0;j<num*2;j++)
    {
    if(str[i][j]=="#")
    System.out.print("#");
    else if(str[i][j]=="/")
    {
    System.out.print("\\");
    }
    else
    {
    System.out.print("/");
    }
    }
    System.out.println("");
    }


    }
    }

    ReplyDelete
  4. Anonymous2 August 2017 at 05:49

    uncomplete Code!!

    ReplyDelete

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