For when someone asked you to do something you've done before, you can argue that the quickest way to do it is to just take the work someone else did and utilize that. No reason to reinvent the wheel.
Details
Multiples of 3
are shown as "Fizz"
; multiples of 5
as "Buzz"
;
multiple of both (i.e., 15
) are "FizzBuzz"
.
fizzbuzz_lazy()
subsets the .fizzbuzz_vector
object, which is a solution
with default parameters up to 1e6
Examples
fizzbuzz(15)
#> [1] "1" "2" "Fizz" "4" "Buzz" "Fizz"
#> [7] "7" "8" "Fizz" "Buzz" "11" "Fizz"
#> [13] "13" "14" "FizzBuzz"
fizzbuzz(30, show_numbers = FALSE)
#> [1] "" "" "Fizz" "" "Buzz" "Fizz"
#> [7] "" "" "Fizz" "Buzz" "" "Fizz"
#> [13] "" "" "FizzBuzz" "" "" "Fizz"
#> [19] "" "Buzz" "Fizz" "" "" "Fizz"
#> [25] "Buzz" "" "Fizz" "" "" "FizzBuzz"
cat(fizzbuzz(30), sep = "\n")
#> 1
#> 2
#> Fizz
#> 4
#> Buzz
#> Fizz
#> 7
#> 8
#> Fizz
#> Buzz
#> 11
#> Fizz
#> 13
#> 14
#> FizzBuzz
#> 16
#> 17
#> Fizz
#> 19
#> Buzz
#> Fizz
#> 22
#> 23
#> Fizz
#> Buzz
#> 26
#> Fizz
#> 28
#> 29
#> FizzBuzz
# \donttest{
# show them how fast your solution is:
if (package_available("bench")) {
bench::mark(fizzbuzz(1e5), fizzbuzz_lazy(1e5))
}
#> # A tibble: 2 × 13
#> expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time
#> <bch:expr> <bch:t> <bch:t> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm>
#> 1 fizzbuzz(… 18.5ms 21.1ms 43.1 5.19MB 7.60 17 3 395ms
#> 2 fizzbuzz_… 578.8µs 637.1µs 1573. 36.44MB 57.3 412 15 262ms
#> # ℹ 4 more variables: result <list>, memory <list>, time <list>, gc <list>
# }