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Published at Jul 13 2018
·
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Instructions

Test suite

Solution

Given a string containing brackets `[]`

, braces `{}`

and parentheses `()`

,
verify that all the pairs are matched and nested correctly.

Ginna Baker

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
module Test.Main where
import Prelude
import Effect (Effect)
import Test.Unit.Assert as Assert
import Test.Unit (TestSuite, suite, test)
import Test.Unit.Main (runTest)
import BracketPush (isPaired)
main :: Effect Unit
main = runTest suites
suites :: TestSuite
suites = do
suite "BracketPush.isPaired" do
test "paired square brackets" $
Assert.equal true
(isPaired "[]")
test "empty string" $
Assert.equal true
(isPaired "")
test "unpaired brackets" $
Assert.equal false
(isPaired "[[")
test "wrong ordered brackets" $
Assert.equal false
(isPaired "}{")
test "paired with whitespace" $
Assert.equal true
(isPaired "{ }")
test "simple nested brackets" $
Assert.equal true
(isPaired "{[]}")
test "several paired brackets" $
Assert.equal true
(isPaired "{}[]")
test "paired and nested brackets" $
Assert.equal true
(isPaired "([{}({}[])])")
test "unopened closing brackets" $
Assert.equal false
(isPaired "{[)][]}")
test "unpaired and nested brackets" $
Assert.equal false
(isPaired "([{])")
test "paired and wrong nested brackets" $
Assert.equal false
(isPaired "[({]})")
test "math expression" $
Assert.equal true
(isPaired "(((185 + 223.85) * 15) - 543)/2")
test "complex latex expression" $
Assert.equal true
(isPaired "\\left(\\begin{array}{cc} \\frac{1}{3} & x\\\\ \\mathrm{e}^{x} &... x^2 \\end{array}\\right)")
```

```
module BracketPush
( isPaired
) where
import Prelude
import Data.Array (elem, findIndex, scanl)
import Data.Maybe (Maybe(..), isJust, maybe)
import Data.String (splitAt, toCharArray, uncons)
isPaired β· String β Boolean
isPaired s = case uncons s of
Nothing β -- empty string is paired
true
Just { head, tail } β
if head `elem` closingChars then
false
else
if head `elem` openingChars then
-- we have to find a closing "opposite"
if hasClosingParen head tail then
-- if we found one, everything "inside" has to be paired
-- and also everything after it.
-- e.g. "[inner]rest"
isPaired (inner head tail) && isPaired (rest head tail)
else
false
else
isPaired tail -- nothing to see here, continue
where
openingChars β· Array Char
openingChars = ['[', '{', '(']
closingChars β· Array Char
closingChars = map opposite openingChars
opposite β· Char β Char
opposite '[' = ']'
opposite '{' = '}'
opposite '(' = ')'
opposite c = c
-- would have liked to curry even more, but
-- hasClosingParen = isJust <<< opposingParenIndex does not work (!?)
hasClosingParen β· Char β String β Boolean
hasClosingParen h = isJust <<< opposingParenIndex h
-- search for the index of the opposing character for h in t,
-- taking nesting into account, e.g.
-- opposingParenIndex '[' "[[{}]foobar]baz" β 11
opposingParenIndex β· Char β String β Maybe Int
opposingParenIndex h t =
findIndex (_ == 0) (scanl (parenCount h) 1 $ toCharArray t)
where
parenCount β· Char β Int β Char β Int
parenCount c a ch
| ch == c = a + 1
| ch == (opposite c) = a - 1
| otherwise = a
-- extract the part inside from string consisting of head and tail
-- e.g. inner '[' "foobar]" β "foobar"
inner β· Char β String β String
inner h t =
maybe "" (\r β r.before) $ (\i β splitAt i t) =<< opposingParenIndex h t
-- extract the part after pair from string consisting of head and tail
-- e.g. inner '[' "foobar]baz" β "baz"
rest β· Char β String β String
rest h t =
maybe "" (\r β r.after) $ (\i β splitAt (i+1) t) =<< opposingParenIndex h t
```

A huge amount can be learned from reading other peopleβs code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

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