Rounded Rectangles

Question

Challenge

Given an integer greater or equal to 4, n, print a rounded rectangle of as close as possible (with a gap of 1) sides and a perimeter of n characters.

Rules

• n is always 4 or greater, because otherwise the output wouldn't be a square
• The characters for the perimeter can be any non-whitespace character
• The rectangle should have equal sides when possible
• When not possible, the rectangle can be up to one character taller or wider than its other side
• The rectangle should have one whitespace character in each corner
• The rectangle can only ever have a gap of 1 (excluding corners) along the entire perimeter
• The gap can be on any side and any position along the perimeter

Rules

Least amount of bytes wins!

Examples

Input: 4
Output:
 o
o o
 o

Input: 5
Output:
 oo
o  o
 o

Input: 8
Output:
 oo
o  o
o  o
 oo

Input: 21
Output:
 oooooo
o      o
o      o
o      o
o      o
o      o
 ooooo

Here is the python3 code I used to generate the above examples:

import math

ch = 'o'

def print_rect(p):
    side_length = math.ceil(p / 4)
    height = side_length
    remainder = p - side_length * 3
    p = side_length + 2

    if side_length - remainder >= 2 or not remainder:
        remainder += 2
        height -= 1

    lines = [(ch * side_length).center(p)]
    for i in range(height):
        lines.append(ch + ch.rjust(side_length + 1))
    lines.append((ch * remainder).center(p))

    print('Output:')
    for line in lines:
        print(line)

print_rect(int(input('Input: ')))

Answer 1

Python 2, 91 bytes

def f(n):t=(n+3)/4;c,s='o ';h=(n-t+1)/3;print'\n'.join([s+c*t]+[c+s*t+c]*h+[s+c*(n-t-h-h)])

Try it online!

Answer 2

Charcoal, 24 bytes

Ｆ⁴«×o⁺÷Ｉθ⁴﹪÷﹪Ｉθ⁴Ｘ²↔⊖鲶↷

Try it online! Link is to verbose version of code. Explanation:

Ｆ⁴«

Loop over the four sides of the rounded rectangle.

×o⁺

Print a number of os given by the sum of...

÷Ｉθ⁴

... one quarter of the input as a number, rounded down, and...

﹪÷﹪Ｉθ⁴Ｘ²↔⊖ι²

... add an extra o a) on both horizontal sides if the input has a remainder (modulo 4) of 2 or 3, and b) on the first vertical side if the input is odd.

¶

Omit the corner.

↷

Pivot ready for the next side.

Answer 3

JavaScript, 143

n=>(x=Math.floor(n/4),q=n%2==1?x+1:x,` ${"o".repeat(q)} 
${Array(n%4==2||n%4==3?x+1:x).fill(`o${" ".repeat(q)}o`).join`\n`}
 ${"o".repeat(x)}`)