Python Program to print the pattern ‘G’

In this article, we will learn how to print the pattern G using stars and white-spaces. Given a number n, we will write a program to print the pattern G over n lines or rows.
Examples: 

Input : 7
Output :
  ***  
 *     
 *     
 * *** 
 *   * 
 *   * 
  ***  

Input : 9
Output :
  *****  
 *       
 *       
 *       
 *   *** 
 *     * 
 *     * 
 *     * 
  *****  

In this program, we have used the simple logic of iteration over lines to create the pattern G. Please look at the image below which represents the pattern G in the form of a 2-d matrix, where mat[i][j] = ‘ij’: 

If we try to analyze this picture with a (row, column) matrix and the circles represent the position of stars in the pattern G, we will learn the steps. Here we are performing the operations column-wise. So for the first line of stars, we set the first if condition, where the row position with 0 and (n-1) won’t get the stars and all other rows from 1 to (n-1), will get the stars. Similarly, for the second, third and fourth column we want stars at the position row = 0 and row = (n-1). The other steps are self-explanatory and can be understood from the position of rows and columns in the diagram.
Below is the implementation of above idea: 

Python

# Python program to print pattern G
def Pattern(line):
    pat=""
    for i in range(0,line):
        for j in range(0,line):
            if ((j == 1 and i != 0 and i != line-1) or ((i == 0 or
                i == line-1) and j > 1 and j < line-2) or (i == ((line-1)/2)
                and j > line-5 and j < line-1) or (j == line-2 and
                i != 0 and i != line-1 and i >=((line-1)/2))):
                pat=pat+"*"
            else:
                pat=pat+" "
        pat=pat+"
"
    return pat

# Driver Code
line = 7
print(Pattern(line))

C++

// C++ program to print pattern G

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

string Pattern(int line){
    string pat = "";
    for(int i=0; i<line; i++){
        for(int j=0; j<line; j++){
            if ((j == 1 and i != 0 and i != line-1) or ((i == 0 or
                i == line-1) and j > 1 and j < line-2) or (i == ((line-1)/2)
                and j > line-5 and j < line-1) or (j == line-2 and
                i != 0 and i != line-1 and i >=((line-1)/2))){
                pat=pat+"*";
            } else{
                pat = pat + " ";
            }
        pat = pat + "                                                                                                                                                              ";
        }
    }

    return pat;
}

// Driver Code
int main() {
    int line = 7;
    cout<<Pattern(line);
}

//This code is contributed by Kunal Mali

Output: 

  ***  
 *     
 *     
 * *** 
 *   * 
 *   * 
  ***  

Time Complexity: O(n²), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Program to print the pattern ‘G’ for more details!