CF 188D - Asterisks

Rating: 1100
Tags: *special, implementation
Solve time: 57s
Verified: yes

Solution

Problem Understanding

The problem asks us to generate a simple text pattern. We are given a single integer n, which represents the number of lines we need to print. The first line should contain a single asterisk, the second line two asterisks, and so on, until the n-th line contains n asterisks. Essentially, the output is a left-aligned right triangle of asterisks with height n.

The input constraint is 1 ≤ n ≤ 50. This tells us that n is small and any solution that runs in linear or quadratic time relative to n will execute almost instantly. There is no risk of hitting performance limits here, so we do not need to consider algorithmic optimizations for large input sizes.

A subtle edge case is when n equals 1. The correct output is a single line with one asterisk. A careless solution that uses a loop starting from zero or off-by-one logic could produce zero lines or fail to print the single asterisk. Similarly, n = 50 is the largest input, so the last line should correctly contain exactly 50 asterisks. Any miscount in the loop boundary could either truncate the line or produce an extra asterisk.

Approaches

The naive approach is straightforward: iterate from 1 to n, and for each iteration i, print i asterisks. This works because each line independently depends only on its index, and constructing a string of repeated characters is a constant-time operation in Python for small n. The brute-force solution has a time complexity proportional to the sum 1 + 2 + … + n, which equals n(n+1)/2. For n ≤ 50, this is at most 1275 character prints, which is trivial for modern computers.

Since the problem is already small-scale, no further optimization is necessary. The key insight is recognizing that each line's content is a simple repetition of a single character, which allows a one-liner in Python using the multiplication operator for strings. The problem does not require data structures or recursion.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(n)	Accepted
Optimal	O(n^2)	O(n)	Accepted

Even though both approaches share the same asymptotic complexity, the "optimal" version leverages string repetition and direct printing to make the code concise.

Algorithm Walkthrough

Read the integer n from input. This defines the number of lines in the triangle.
Iterate over the range from 1 to n inclusive. Let the loop variable be i.
For each i, generate a string consisting of i asterisks. In Python, this is done with "*" multiplied by i.
Print the generated string. Each iteration prints one line, automatically appending a newline.

Why it works: The loop invariant is that on iteration i, exactly i asterisks are printed. This guarantees that after the final iteration, the triangle has the correct shape and each line contains the exact number of asterisks corresponding to its 1-based index.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())

for i in range(1, n + 1):
    print('*' * i)

The code starts by reading the input using fast I/O. Converting the input string to an integer ensures we have a proper loop bound. The for loop iterates over each line index from 1 to n, generating a string with the correct number of asterisks and printing it immediately. Multiplying a string by an integer in Python produces a new string with repeated characters, which matches the required pattern. Off-by-one errors are avoided by using range(1, n + 1) instead of range(n).

Worked Examples

Sample Input 1:

i	'' i	Printed line
1	'*'	*
2	'**'	**
3	'***'	***

This confirms that the algorithm correctly prints one asterisk on the first line, two on the second, and three on the third.

Sample Input 2:

i	'' i	Printed line
1	'*'	*
2	'**'	**
3	'***'	***
4	'****'	****
5	'*****'	*****

This demonstrates that the solution scales linearly with the number of lines and respects the 1-based line indexing.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n^2)	The sum of characters printed is 1 + 2 + … + n = n(n+1)/2, which is O(n^2).
Space	O(n)	Each line requires storing up to n characters before printing.

Given that n ≤ 50, the algorithm executes instantly and uses negligible memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    n = int(input())
    for i in range(1, n + 1):
        print('*' * i)
    return output.getvalue().strip()

# provided sample
assert run("3\n") == "*\n**\n***", "sample 1"

# custom cases
assert run("1\n") == "*", "minimum input"
assert run("50\n") == '\n'.join('*' * i for i in range(1, 51)), "maximum input"
assert run("2\n") == "*\n**", "small even number"
assert run("5\n") == "*\n**\n***\n****\n*****", "small odd number"

Test input	Expected output	What it validates
1	*	Handles minimum input
50	1 to 50 asterisks per line	Correctly prints maximum allowed lines
2	, *	Correct line count for small even input
5	, , , , ***	Correct line count for small odd input

Edge Cases

When n = 1, the algorithm runs a single iteration, printing exactly one asterisk. This confirms the code correctly handles the smallest input. For n = 50, the loop runs 50 times, each line containing the correct number of asterisks. Multiplying the string by the loop variable ensures no off-by-one errors occur. All lines are printed sequentially, preserving the required ordering.