Webkul Interview Coding Round Pattern

Difficulty Level : Hard
Last Updated : 10 Dec, 2020

This is a mixed pattern which uses the different types in one and this pattern was asked to me when I was giving the 1st round of coding in Webkul Interview.

Examples:

Input : 3
Output : 
                            @
                           @@@
                          @@@@@
                          *   *
                         **@@@**
                          *   *

Input : 5
Output : 
                            @
                           @@@
                          @@@@@
                         @@@@@@@
                         *     *
                        **     **
                       ***@@@@@***
                        **     **
                         *     *

Below is the implementation of the above pattern:

C++

// C++ implementation
#include <bits/stdc++.h>
  
using namespace std;
 
// Driver code
int main() {
    int n, i, j;
    cin >> n ;
     
    for (i = 0; i < (n/2) + 2; i++){
        for (j = 0; j < n - i; j++){
            cout << " ";
        }
        for (j = 0; j < 2 * i + 1; j++){
            cout <<"@";
        }
        cout << "\n";
    }
     
    for (i = 0; i < n; i++){
        for (j = 0; j < (n/2) + 1; j++){
            if (j >= n / 2 - i && j >= i - n / 2){
                cout << "*" ;
            }
            else
            cout << " ";
        }
        for (j = 0; j < n; j++){
            if (i == n / 2)
                cout <<"@" ;
            else
                cout <<" ";
        }
        for (j = 0; j < (n / 2) + 1; j++){
            if (j >= n / 2 - i && j >= i - n / 2)
                cout <<"*";
        }
        cout <<"\n";
    }
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

// C implementation
#include <stdio.h>
 
// Driver code
int main() {
    int n, i, j;
    scanf("%d", &n);
     
    for (i = 0; i < (n/2) + 2; i++){
        for (j = 0; j < n - i; j++){
            printf(" ");
        }
        for (j = 0; j < 2 * i + 1; j++){
            printf("@");
        }
        printf("\n");
    }
     
    for (i = 0; i < n; i++){
        for (j = 0; j < (n/2) + 1; j++){
            if (j >= n / 2 - i && j >= i - n / 2){
                printf("*");
            }
            else
            printf(" ");
        }
        for (j = 0; j < n; j++){
            if (i == n / 2)
                printf("@");
            else
                printf(" ");
        }
        for (j = 0; j < (n / 2) + 1; j++){
            if (j >= n / 2 - i && j >= i - n / 2)
                printf("*");
        }
        printf("\n");
    }
    return 0;
}
 
// This code is contributed by shubhamsingh10

Java

// Java implementation
import java.util.*;
 
class GFG{
     
// Driver Code   
public static void main(String[] args)
{
    int n, i, j;
    Scanner s = new Scanner(System.in);
 
    n = s.nextInt();
       
    for(i = 0; i < (n / 2) + 2; i++)
    {
        for(j = 0; j < n - i; j++)
        {
            System.out.print(" ");
        }
        for(j = 0; j < 2 * i + 1; j++)
        {
            System.out.print("@"); 
        }
        System.out.println();
    }
       
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < (n / 2) + 1; j++)
        {
            if (j >= n / 2 - i && j >= i - n / 2)
            {
                System.out.print("*");
            }
            else
            System.out.print(" ");
        }
        for(j = 0; j < n; j++)
        {
            if (i == n / 2)
                System.out.print("@");
            else
                System.out.print(" ");
        }
        for(j = 0; j < (n / 2) + 1; j++)
        {
            if (j >= n / 2 - i && j >= i - n / 2)
                System.out.print("*");
        }
        System.out.println();
    }
}
}
 
// This code is contributed by divyesh072019

Python3

# Python3 implementation
n=int(input())
 
for i in range(n//2+2):
    for j in range(n-i):
        print(" ",end="")
    for j in range(2*i+1):
        print("@",end="")
    print()
 
for i in range(n):
    for j in range(n//2+1):
        if (j>=n//2-i and j>=i-n//2):
            print("*",end="")
        else:
            print(" ",end="")
    for j in range(n):
        if i==n//2:
            print("@",end="")
        else:
            print(" ",end="")
    for j in range(n//2+1):
        if (j>=n//2-i and j>=i-n//2):
            print("*",end="")
    print()

C#

// C# implementation
using System;
class GFG {
  static void Main()
  {
       
    int n, i, j;
    string val;
    val = Console.ReadLine();
    n = Convert.ToInt32(val);
      
    for (i = 0; i < (n/2) + 2; i++)
    {
        for (j = 0; j < n - i; j++)
        {
            Console.Write(" ");
        }
        for (j = 0; j < 2 * i + 1; j++)
        {
            Console.Write("@");
        }
        Console.WriteLine();
    }
      
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < (n/2) + 1; j++)
        {
            if (j >= n / 2 - i && j >= i - n / 2)
            {
                Console.Write("*");
            }
            else
            Console.Write(" ");
        }
        for (j = 0; j < n; j++)
        {
            if (i == n / 2)
                Console.Write("@");
            else
                Console.Write(" ");
        }
        for (j = 0; j < (n / 2) + 1; j++)
        {
            if (j >= n / 2 - i && j >= i - n / 2)
                Console.Write("*");
        }
        Console.WriteLine();
    }
  }
}
 
// This code is contributed by divyeshrabadiya07

Input:

Output:

     @
    @@@
   @@@@@
  @@@@@@@
  *     *
 **     **
***@@@@@***
 **     **
  *     *

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My Personal Notes

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Webkul Interview Coding Round Pattern

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C#