In the first part, I had this problem to fix:
The trick is to find out how to go from a "number of spaces to distribute" to a "special spaces array".
I worked through a few examples to get an intuitive feel of how the logic could look like:
Suppose there are 7 spaces to distribute in 3 slots:
spaces to distribute -> 7
slots -> 3
I could first distribute equally by dividing the two (integer division):
equal division -> 7 / 3 = 2 (remainder 1)
-> ["--", "--", "--"]
And then distribute the remainder
left-to-right:
equal division -> 7 / 3 = 2 (remainder 1)
-> ["--", "--", "--"]
with remainder -> ["---", "--", "--"]
distributed
I tried to run some more examples for this. Like this:
spaces to distribute -> 15
slots to distribute -> 4
equal distribution -> 15/4 => 3
-> ["---","---","---","---"]
remainder -> 3
remainder distributed -> ["----","----","----","---"]
This still did not give me a clear grasp of the distribution logic.
Eventually, I came up with this idea: I could have a function that gradually "builds" the array of spaces from an empty array.
It would do this by "remembering" and "updating" the number of extra/remainder spaces it has to distribute across the slots.
So, something like this:
example: 15 spaces, 4 slots
step 1: empty array ([]), number of slots to fill (4), number of remainders to distribute (remainder 15/4 = 3)
step 2:
- push the first space slot into the array [?]
- where ? = normally, just the equal distribution of space.
- in this case, that's 15/4 quotient = 3. so => [3]
- but the number of remainders to distribute is more than 0,
- so we just add 1 more to the fill => [4]
- now that we've filled the first, we have to reduce number of slots to fill to 3.
- also, we've distributed 1 of the 3 remainders to distribute so that also becomes 2.
step 3:
- repeat the above step for the next slot,
but remember that we only have 2 remainders to distribute and 3 slots to fill.
...and so on...
Doing this "recursively" gives us a final array of space slots.
The function looks something like this:
import Prelude
import Data.Array (snoc, catMaybes)
import Data.Int (quot, rem)
import Data.String.Utils (repeat)
makeSpaces :: Int -> ValidLine -> Array String
makeSpaces maxLen vl =
let
d = deficit maxLen vl
spaceSlots = numberOfSpaces vl
totalSpaces = d + spaceSlots
in
makeSpacesHelper [] (quot totalSpaces spaceSlots) (rem totalSpaces spaceSlots) spaceSlots
# map (\a -> repeat a "-")
-- note: I'm using hyphens for demonstration
-- use space in the real program
# catMaybes
makeSpacesHelper :: Array Int -> Int -> Int -> Int -> Array Int
makeSpacesHelper xs _ _ 0 = xs
makeSpacesHelper xs n incCount timesCount =
makeSpacesHelper (snoc xs a) n (incCount-1) (timesCount-1)
where
a = n + (if incCount > 0 then 1 else 0)
In this:
incCount => the remainders to be distributed
timesCount => number of slots to be filled
I've used a bunch of helper functions in the above code like deficit
, numberOfSpaces
and some library functions like catMaybes
.
Here's definitions for the helpers:
import Data.Array (length, foldl)
import Data.String as Str
deficit :: Int -> ValidLine -> Int
deficit maxLen vl = maxLen - (totalCharLength vl)
numberOfSpaces :: ValidLine -> Int
numberOfSpaces (ValidLine xs) = length xs - 1
totalCharLength :: ValidLine -> Int
totalCharLength (ValidLine xs) =
foldl (\b a -> b + Str.length a) 0 xs + numberOfSpaces (ValidLine xs)
Testing all of this:
> makeSpaces 16 (ValidLine ["this","is","an"])
# that is max length is 16. valid line has char length = 10.
# deficit = 6. Total spaces to distribute is 6+2 = 8. Across 2 slots.
["----","----"]
> makeSpaces 19 (ValidLine ["a","b","c","d","e"])
# that is, max length is 19, valid line has 9 characters length, so deficit is 10.
# Total spaces to distribute is 10+4 = 14. Across 4 slots.
["----","----","---","---"]
All we need now is a way to mix an array of strings (from the ValidLine
) and this spaces array such that the words and spaces are interleaved.
I checked Purescript's Array package and found a transpose
function that fit the need perfectly:
transpose :: forall a. Array (Array a) -> Array (Array a)
{-
The 'transpose' function transposes the rows and columns of its argument. For example,
transpose
[ [1, 2, 3]
, [4, 5, 6]
] ==
[ [1, 4]
, [2, 5]
, [3, 6]
]
If some of the rows are shorter than the following rows, their elements are skipped:
transpose
[ [10, 11]
, [20]
, [30, 31, 32]
] ==
[ [10, 20, 30]
, [11, 31]
, [32]
]
-}
So I could just put the array of valid lines and array of space slots into another array and then transpose
them.
import Data.Array (transpose, concat)
specialIntercalate :: forall a. Array a -> Array a -> Array a
specialIntercalate xs ys = transpose [xs, ys] # concat
Mixing these so far:
validLineToParaLine :: Int -> ValidLine -> String
validLineToParaLine maxLen (ValidLine xs) =
specialIntercalate xs (makeSpaces maxLen (ValidLine xs))
# Str.joinWith ""
And the test:
> validLineToParaLine 19 (ValidLine ["a","b","c","d","e"])
"a----b----c---d---e"
> validLineToParaLine 16 (ValidLine ["This","is","an"])
"This----is----an"
All that remains now is getting ValidLine
s out of the given array of words.
I'll write about that in the next part.
Top comments (0)