Ruby Program to Print Star Pattern

You can print a star pattern in Ruby using nested loops like 5.times { |i| puts '*' * (i + 1) } which prints stars increasing from 1 to 5 in each line.

Examples

Input5

Output* ** *** **** *****

Input3

Output* ** ***

Input1

Output*

How to Think About It

To print a star pattern, think of each line as having a number of stars equal to the line number. Start from line 1 with one star, then line 2 with two stars, and so on until the given number of lines. Use a loop to repeat this printing for each line.

Algorithm

1

Get the number of lines to print from the user or set a fixed number.

2

Start a loop from 1 to the number of lines.

3

For each iteration, print stars equal to the current line number.

4

Move to the next line and repeat until all lines are printed.

Code

ruby

puts 'Enter number of lines:'
lines = gets.to_i

(1..lines).each do |i|
  puts '*' * i
end

Output

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

Dry Run

Let's trace the example where the input is 5 lines through the code.

1

Input number of lines

User inputs 5, so lines = 5

2

Start loop from 1 to 5

Loop variable i starts at 1

3

Print stars for each line

For i=1, print '*' * 1 => '*'

4

Next iterations

i=2 prints '**', i=3 prints '***', i=4 prints '****', i=5 prints '*****'

i	Stars Printed
1	*
2	**
3	***
4	****
5	*****

Why This Works

Step 1: Loop controls lines

The loop runs from 1 to the number of lines, controlling how many lines are printed.

Step 2: Stars printed per line

For each line number i, printing '*' repeated i times creates the star pattern.

Step 3: Output formatting

Each puts prints stars and moves to the next line automatically.

Alternative Approaches

Using times loop

ruby

puts 'Enter number of lines:'
lines = gets.to_i

lines.times do |i|
  puts '*' * (i + 1)
end

This uses zero-based index from times, so add 1 to get correct star count.

Using while loop

ruby

puts 'Enter number of lines:'
lines = gets.to_i
count = 1
while count <= lines
  puts '*' * count
  count += 1
end

While loop is more manual but works the same way.

Complexity: O(n^2) time, O(1) space

Time Complexity

The outer loop runs n times, and each line prints up to n stars, so total operations are proportional to n squared.

Space Complexity

No extra space is used except for loop variables and temporary strings, so space is constant.

Which Approach is Fastest?

All approaches have similar time and space complexity; using times or each is mostly a style choice.

Approach	Time	Space	Best For
each loop	O(n^2)	O(1)	Readability and clarity
times loop	O(n^2)	O(1)	Concise code with zero-based index
while loop	O(n^2)	O(1)	Manual control, beginner-friendly