CF 102697109 - H.

The task is a small output-construction problem. We are given one integer n, which specifies the size of every part of a capital letter H. Each vertical leg contains n characters, and the horizontal bar has the same length as the distance between the two legs.

codeforcescompetitive-programming

CF 102697109 - H.

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Solve time: 1m 24s
Verified: yes

Solution

Problem Understanding

The task is a small output-construction problem. We are given one integer n, which specifies the size of every part of a capital letter H. Each vertical leg contains n characters, and the horizontal bar has the same length as the distance between the two legs. The required drawing has n rows above the horizontal bar, one middle row containing the entire bar, and n rows below it.

For example, when n = 3, the two vertical legs occupy columns 0 and 4, while the middle bar occupies all five columns. Thus the final drawing has 2n + 1 rows and 2n - 1 columns. The output is simply this rectangular character pattern, with H at the two outer columns on non-middle rows and H in every position of the middle row.

The statement gives no explicit upper bound for n in the available version. The relevant constraint is consequently the amount of output itself. A drawing contains (2n + 1)(2n - 1) = 4n² - 1 character positions before newline characters are counted, so no solution can asymptotically use less than O(n²) time if it explicitly prints the drawing. The one-second limit also means there is no reason to perform anything more expensive than directly constructing the required rows.

The main edge case is n = 1. In that case the two legs and the middle bar all coincide, so the answer is three rows containing a single H, not a shape with two separated columns. The exact input and output are:

1

H
H
H

A careless implementation that always assumes the two legs occupy different columns could accidentally create extra spaces or an empty gap.

Another boundary case is the middle row. For n = 2, the output is:

H H
HHH
H H

The middle row must contain 2n - 1 = 3 consecutive H characters. An implementation that prints the vertical-leg pattern for every row would produce H H in the center and miss the horizontal bar entirely.

Approaches

A direct brute-force approach is to consider every position in the output rectangle. For a position (r, c), print H when r is the middle row or when c is one of the two outer columns. Otherwise print a space. Since the output has exactly (2n + 1)(2n - 1) = 4n² - 1 positions, this performs exactly 4n² - 1 position checks. It is correct because the definition of the letter H is completely characterized by those three lines: the left leg, the right leg, and the horizontal middle bar.

The brute-force version is already asymptotically optimal in time because the output itself has quadratic size. There is no hidden algorithmic challenge that allows us to avoid producing those characters. The useful improvement is instead to avoid storing the entire drawing. We can construct one row at a time and immediately append it to the output. A non-middle row is simply H followed by 2n - 3 spaces and another H, while the middle row consists of 2n - 1 H characters.

This also handles n = 1 naturally if the row construction is chosen carefully. In that case there is only one column, so the non-middle-row formula should not attempt to produce two separate legs. The simplest implementation is to generate the first n rows as the vertical pattern, generate the middle row, then generate the remaining n rows. When n = 1, the vertical pattern is just H, and the middle row is also H.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n²) O(1) extra Accepted
Optimal O(n²) O(n) output buffer Accepted

The two approaches have the same asymptotic time because printing dominates the problem. The optimal implementation is cleaner because it constructs complete rows directly instead of checking every individual cell.

Algorithm Walkthrough

  1. Read n. The final drawing has 2n + 1 rows and 2n - 1 columns, so these dimensions completely determine the output.
  2. Construct the pattern for a non-middle row. If n = 1, the row is simply H. Otherwise it consists of H, followed by 2n - 3 spaces, followed by another H. The two H characters are exactly the two vertical legs.
  3. Output this vertical row n times. These rows form the upper half of the letter and contain no horizontal bar.
  4. Output a row consisting of 2n - 1 consecutive H characters. This is the horizontal bar and is exactly one row wide.
  5. Output the same vertical row another n times. These rows form the lower half of the letter.

The invariant throughout the construction is that every emitted row is exactly the row required at its position. The first and last n rows contain only the two vertical legs, while the unique middle row contains the complete horizontal bar. Since these are all 2n + 1 rows, every position in the required drawing is produced exactly once and with the correct character.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())

    if n == 1:
        row = "H"
    else:
        row = "H" + " " * (2 * n - 3) + "H"

    middle = "H" * (2 * n - 1)

    out = []
    out.extend([row] * n)
    out.append(middle)
    out.extend([row] * n)

    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()

The first branch handles the only degenerate dimension. For n > 1, the distance between the two vertical H characters is 2n - 3 spaces. Including the two endpoint characters gives a total width of 2 + (2n - 3) = 2n - 1.

The list out stores complete rows rather than individual characters. The first n entries represent the upper legs, the next entry is the middle bar, and the final n entries represent the lower legs. Using "\n".join(out) avoids adding an unnecessary trailing newline and performs the output in one operation.

There is no integer overflow concern in Python, and there are no index calculations involving the input array. The only arithmetic that needs care is 2n - 3, which is valid for the separate n = 1 case handled above.

Worked Examples

Sample 1

For n = 3, the drawing has 7 rows and 5 columns. The vertical row is H H, because there are 2 * 3 - 3 = 3 spaces between the legs. The middle row is HHHHH.

Step Rows generated Current row
1 1 H H
2 2 H H
3 3 H H
4 4 HHHHH
5 5 H H
6 6 H H
7 7 H H

The result is:

H   H
H   H
H   H
HHHHH
H   H
H   H
H   H

This demonstrates the general case. There are exactly three rows on either side of the horizontal bar, and the bar spans the complete width.

Sample 2

The statement provides only one sample, so a second trace can use n = 2. Here the drawing width is 3, and a non-middle row is H H.

Step Rows generated Current row
1 1 H H
2 2 H H
3 3 HHH
4 4 H H
5 5 H H

The result is:

H H
H H
HHH
H H
H H

The trace confirms that the horizontal bar is placed after exactly n upper rows and that the total number of rows is 2n + 1.

Complexity Analysis

Measure Complexity Explanation
Time O(n²) The program produces 2n + 1 rows of width 2n - 1, so the amount of printed data is Θ(n²).
Space O(n²) The out list stores the complete output before writing it.

The quadratic time is unavoidable because the required output itself contains Θ(n²) characters. The implementation performs only constant work per output character apart from string construction, so there is no algorithmic overhead beyond the required printing.

If strict streaming output were desired, the same construction could print each row immediately and reduce auxiliary memory to O(n), since a single row has length O(n). The submitted version keeps complete rows in memory to make the output operation simple and efficient.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    n = int(input())

    if n == 1:
        row = "H"
    else:
        row = "H" + " " * (2 * n - 3) + "H"

    middle = "H" * (2 * n - 1)

    out = []
    out.extend([row] * n)
    out.append(middle)
    out.extend([row] * n)

    sys.stdout.write("\n".join(out))

def run(inp: str) -> str:
    global input

    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    input = sys.stdin.readline
    solve()

    result = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout

    return result

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

# minimum-size input
assert run("1\n") == (
    "H\n"
    "H\n"
    "H"
), "minimum size"

# small boundary case
assert run("2\n") == (
    "H H\n"
    "H H\n"
    "HHH\n"
    "H H\n"
    "H H"
), "n = 2"

# larger custom case
assert run("4\n") == (
    "H     H\n"
    "H     H\n"
    "H     H\n"
    "H     H\n"
    "HHHHHHH\n"
    "H     H\n"
    "H     H\n"
    "H     H\n"
    "H     H"
), "n = 4"

# stress-style case, also checks that the formula scales correctly
result = run("1000\n")
lines = result.splitlines()
assert len(lines) == 2001, "row count"
assert all(len(line) == 1999 for line in lines), "row width"
assert lines[999] == "H" * 1999, "middle row"
assert lines[0] == "H" + " " * 1997 + "H", "upper leg"
assert lines[-1] == "H" + " " * 1997 + "H", "lower leg"
Test input Expected output What it validates
1 Three lines containing H Degenerate case where both legs coincide
2 A 5 × 3 H Boundary arithmetic for the smallest non-degenerate shape
4 A 9 × 7 H Correct spacing and row counts in a general case
1000 2001 rows of width 1999 Large-output behavior and off-by-one boundaries

Edge Cases

For n = 1, the input is:

1

The algorithm chooses row = "H" instead of applying the general two-leg formula. It then outputs that row once above the middle row, once as the middle row, and once below it. The result is:

H
H
H

This avoids the common mistake of computing a negative or zero-width gap between two supposedly distinct legs.

For n = 2, the input is:

2

The vertical row becomes H H, because 2n - 3 = 1. The middle row becomes HHH, because 2n - 1 = 3. The algorithm emits two vertical rows, one middle row, and two more vertical rows, giving:

H H
H H
HHH
H H
H H

This catches both common off-by-one errors: using 2n - 2 instead of 2n - 3 for the inner spaces, and using 2n instead of 2n - 1 for the middle bar.

For a large value such as n = 1000, the output has 2001 rows and every row has width 1999. The middle row contains exactly 1999 H characters, while every other row contains exactly two H characters at its boundaries. Since the algorithm derives both dimensions directly from n, the same reasoning applies without any special handling for large values.

The key boundary condition is always the distinction between the middle row and every other row. There is exactly one middle row, at position n when rows are counted from zero. The implementation avoids explicit row indexing by emitting exactly n vertical rows, then one middle row, then exactly n more vertical rows, which makes the row count and the symmetry explicit.