Write a function that takes an integer and returns an array `[A, B, C]`, where `A` is the number of multiples of 3 (but not 5) below the given integer, `B` is the number of multiples of 5 (but not 3) below the given integer and `C` is the number of multiples of 3 and 5 below the given integer. 

For example, `solution(20)` should return `[5, 2, 1]`

~~~if:r
```r
# in R, returns a numeric vector
solution(20)
[1] 5 2 1

class(solution(20))
[1] "numeric"
```
~~~

Difficulty: introductory

Starter code:

def solution(number):
	

Solutions:
def solution(number):
    A = (number - 1) // 3
    B = (number - 1) // 5
    C = (number - 1) // 15    
    return [A - C, B - C, C]

def solution(number):
    n = number - 1
    a, b, c = n // 3, n // 5, n // 15
    return [a - c, b - c, c]

def solution(n):
  n -= 1
  return [n//3-n//15, n//5-n//15, n//15]

def solution(number):
    number = number - 1
    fizzbuzz = number // 15
    fizz = number // 3 - fizzbuzz
    buzz = number // 5 - fizzbuzz
    return [fizz, buzz, fizzbuzz]

def solution(number):
    [threes, fives, fifteens] = [int((number-1)/div) for div in [3, 5, 15]]
    return [threes-fifteens, fives-fifteens, fifteens]


def solution(number):
    returnArray = [0, 0, 0]
    for i in range(1, number):
        if i % 15 == 0:
            returnArray[2] += 1
        elif i % 5 == 0 and (not (i % 3 == 0)):
            returnArray[1] += 1
        elif i % 3 == 0:
            returnArray[0] += 1
    return returnArray

def solution(number):
    number = number - 1
    c = int(number/15)
    b = int(number/5) - c
    a = int(number/3) - c
    return [a, b, c]

def solution(number):
    count = 0
    lst = []
    for i in range(1,number):
        if i % 3 == 0 and not i % 5 == 0:
            count += 1
    lst.append(count)
    count = 0
    for i in range(1,number):
        if not i % 3 == 0 and i % 5 == 0:
            count += 1
    lst.append(count)
    count = 0
    for i in range(1,number):
        if i % 3 == 0 and i % 5 == 0:
            count += 1
    lst.append(count)        
    return lst

def solution(n): c = lambda x: (n - 1) // x if x == 15 else (n - 1) // x - c(15); return [c(3), c(5), c(15)]