Pattern 19: Double Triangle Pattern

Summary

This problem is about generating a double triangle pattern using asterisks (*) and spaces. The goal is to create a pattern that resembles two triangles, one inverted and one upright, with increasing sizes.

Detailed Explanation

To solve this problem, we will use a simple algorithmic approach. We will iterate through each row of the pattern, printing the necessary number of spaces and asterisks to form the triangle. We will start by printing the first row of the upright triangle, then the first row of the inverted triangle, followed by the second row of the upright triangle, and so on.

Here's a step-by-step breakdown of the solution:

Initialize an integer variable n to represent the size of the pattern.
Iterate through each row of the pattern:
- For each row, print n-1 spaces (to indent the triangle) followed by 2*(row_number)+1 asterisks (to form the triangle).
Print a newline character (\n) after each row to separate the triangles.
Repeat step 2 until the desired size of the pattern is reached.

Here's an ASCII art diagram illustrating the double triangle pattern:

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

Time complexity analysis: O(n), where n is the size of the pattern. Space complexity analysis: O(1), as we only use a constant amount of space to store variables.

Optimized Solutions

Java

public class DoubleTrianglePattern { public static void main(String[] args) { int n = 5; // adjust this value for different sizes for (int i = 0; i < n; i++) { for (int j = 0; j < n - i - 1; j++) { System.out.print(" "); } for (int k = 0; k < 2 * i + 1; k++) { System.out.print("*"); } System.out.println(); } for (int i = 0; i < n; i++) { for (int j = 0; j < i + 1; j++) { System.out.print(" "); } for (int k = 0; k < 2 * (n - i) - 1; k++) { System.out.print("*"); } System.out.println(); } } }