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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 Data.List (List(..), filter, foldr, fromFoldable, null, (:))
import Data.String (toCharArray)
import Prelude ((>>>))
isPaired :: String -> Boolean
isPaired = toCharArray
>>> fromFoldable
>>> filter isParenthesis
>>> reduceParentheses
>>> null
isParenthesis :: Char -> Boolean
isParenthesis '(' = true
isParenthesis '[' = true
isParenthesis '{' = true
isParenthesis ')' = true
isParenthesis ']' = true
isParenthesis '}' = true
isParenthesis _ = false
reduceParentheses :: List Char -> List Char
reduceParentheses = foldr reduceParenthesis Nil
reduceParenthesis :: Char -> List Char -> List Char
reduceParenthesis '(' (')' : a) = a
reduceParenthesis '[' (']' : a) = a
reduceParenthesis '{' ('}' : a) = a
reduceParenthesis c a = c : a
```

A huge amount can be learned from reading other peopleâ€™s 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?
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