Program to print the pattern 'G' Last Updated : 20 Feb, 2023 Summarize Suggest changes Share Like Article Like Report 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: C++ // C++ program to print the pattern G #include <iostream> using namespace std; void pattern(int line) { int i, j; for(i = 0; i < line; i++) { for(j = 0; j < line; j++) { if((j == 1 && i != 0 && i != line - 1) || ((i == 0 || i == line - 1) && j > 1 && j < line - 2) || (i == ((line - 1) / 2) && j > 2 && j < line - 1) || (j == line - 2 && i != 0 && i >= ((line - 1) / 2) && i != line - 1)) printf("*"); else printf( " "); } printf("\n"); } } // Driver code int main() { int line = 7; pattern(line); return 0; } // This code is contributed // by vt_m. Java // Java program to print the pattern G import java.io.*; class GFG { static void pattern(int line) { int i, j; for(i = 0; i < line; i++) { for(j = 0; j < line; j++) { if((j == 1 && i != 0 && i != line - 1) || ((i == 0 || i == line - 1) && j > 1 && j < line - 2) || (i == ((line - 1) / 2) && j > 2 && j < line - 1) || (j == line - 2 && i != 0 && i >= ((line - 1) / 2) && i != line - 1)) System.out.print("*"); else System.out.print( " "); } System.out.println(); } } // Driver code public static void main (String[] args) { int line = 7; pattern(line); } } // This code is contributed by vt_m. 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+"\n" return pat # Driver Code line = 7 print(Pattern(line)) C# // C# program to print the pattern G using System; class GFG { static void pattern(int line) { int i, j; for(i = 0; i < line; i++) { for(j = 0; j < line; j++) { if((j == 1 && i != 0 && i != line - 1) || ((i == 0 || i == line - 1) && j > 1 && j < line - 2) || (i == ((line - 1) / 2) && j > 2 && j < line - 1) || (j == line - 2 && i != 0 && i >= ((line - 1) / 2) && i != line - 1)) Console.Write("*"); else Console.Write( " "); } Console.WriteLine(); } } // Driver code public static void Main () { int line = 7; pattern(line); } } // This code is contributed by vt_m. PHP <?php // PHP program to print pattern G function Pattern($line){ for ($i=0; $i<$line; $i++) { for ($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 > 2 and $j < $line-1) or ($j == $line-2 and $i != 0 and $i >=(($line-1)/2) and $i != $line-1)) echo "*"; else echo " "; } echo "\n"; } } // Driver Code $line = 7; Pattern($line) ?> JavaScript <script> // JavaScript program to print the pattern G function pattern(line) { var i, j; for (i = 0; i < line; i++) { for (j = 0; j < line; j++) { if ( (j == 1 && i != 0 && i != line - 1) || ((i == 0 || i == line - 1) && j > 1 && j < line - 2) || (i == (line - 1) / 2 && j > 2 && j < line - 1) || (j == line - 2 && i != 0 && i >= (line - 1) / 2 && i != line - 1) ) document.write("*"); else document.write(" "); } document.write("<br>"); } } // Driver code var line = 7; pattern(line); // This code is contributed by rdtank. </script> 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. 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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, 6 min read Python | Print an Inverted Star PatternAn inverted star pattern involves printing a series of lines, each consisting of stars (*) that are arranged in a decreasing order. Here we are going to print inverted star patterns of desired sizes in Python Examples: 1) Below is the inverted star pattern of size n=5 (Because there are 5 horizontal 2 min read Python 3 | Program to print double sided stair-case patternBelow mentioned is the Python 3 program to print the double-sided staircase pattern. Examples: Input: 10Output: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Note: This code only works for even values of n. 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