Daily Challenge #281 - Area or Perimeter

You are given the length and width of a 4-sided polygon. The polygon can either be a rectangle or a square.

If it is a square, return its area. If it is a rectangle, return its perimeter.

area_or_perimeter(6, 10) --> 32
area_or_perimeter(4, 4) --> 16

Tests

area_or_perimeter(5, 5)
area_or_perimeter(10, 20)

Good luck!

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Comments

[Mihail Malo]

        mov     rcx, rsi
        imul    rcx, rdi
        lea     rax, [rsi + rdi]
        add     rax, rax
        cmp     rdi, rsi
        cmove   rax, rcx
        ret

ooga booga

[Deepak Raj]

def area_or_perimeter(a, b):
    if a == b:
        return a * b
    else:
        return 2 * (a + b)

assert area_or_perimeter(5, 5) == 25
assert area_or_perimeter(10, 20) == 60

[Rafael Acioly]

return a * b if a == b else 2 * (a + b)

:)

[peter279k]

Here is the simple solution with Python:

def area_or_perimeter(l , w):
    if l == w:
        return l * w
    return (l + w) * 2

[cromatikap]

In js: let area_or_perimeter = (x, y) => x === y ? x*y : 2*(x+y);

Pass all the tests.

[Hridayesh Sharma]

const area_or_perimeter = (a,b) => a===b ? a*b : 2*(a+b);

[Gagan]

def area_or_perimeter(l,w):
    return l*w if l==w else 2*(l+w)

[Alexander Girke]

In Kotlin it would be:

fun area_or_perimeter(a: Int, b: Int) =
        if (a == b)
            a * b
        else
            2 * (a + b)

[Hasobi]

It's quite simple, check whether a=b if it's true return a*b if it's not, return (a+b)*2.