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alech's solution

to Bracket Push in the PureScript Track

Published at Jul 13 2018 · 0 comments
Test suite

Given a string containing brackets [], braces {} and parentheses (), verify that all the pairs are matched and nested correctly.


Ginna Baker

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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 ∷ String β†’ Boolean
isPaired s = case uncons s of
    Nothing β†’ -- empty string is paired
    Just { head, tail } β†’
        if head `elem` closingChars then
            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)
                isPaired tail -- nothing to see here, continue
        openingChars ∷ Array Char
        openingChars = ['[', '{', '(']

        closingChars ∷ Array Char
        closingChars = map opposite openingChars

        opposite ∷ Char β†’ Char
        opposite '[' = ']'
        opposite '{' = '}'
        opposite '(' = ')'
        opposite c = c

        -- would have liked to curry even more, but
        -- hasClosingParen = isJust <<< opposingParenIndex does not work (!?)
        hasClosingParen ∷ 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 ∷ Char β†’ String β†’ Maybe Int
        opposingParenIndex h t =
            findIndex (_ == 0) (scanl (parenCount h) 1 $ toCharArray t)
                parenCount ∷ 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 ∷ 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 ∷ Char β†’ String β†’ String
        rest h t =
            maybe "" (\r β†’ r.after) $ (\i β†’ splitAt (i+1) t) =<< opposingParenIndex h t

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