Webkul Interview Coding Round Pattern

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 

Python

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() 

Input:

Output:

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

My Personal Notes

C++

C

Python

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