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Homework: Brackets Check in JS

I recentliy saw a homework for students in which you should write a function that gives true if a string with brackets of different types closes brackets accordingly otherwise false.

My first solution not using a stack was this:

const openBrackets = '([{'
const closingBrackets = ')]}'

function checkBrackets (string) {
  let bracketsStack = ''
  for (let i = 0; i < string.length; i++) {
    const char = string[i]
    if (openBrackets.includes(char)) {
      bracketsStack += char
    } else if (closingBrackets.includes(char)) {
      const lastBracket = bracketsStack[bracketsStack.length - 1]
      if ((lastBracket === '(' && char !== ')') ||
          (lastBracket === '{' && char !== '}') ||
          (lastBracket === '[' && char !== ']')) {
        return false
      }
      bracketsStack = bracketsStack.substring(0, bracketsStack.length - 1)
    }
  }
  return true
}

console.log(checkBrackets('()()')) // true
console.log(checkBrackets('([])')) // true
console.log(checkBrackets('([)]')) // false
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But it looks a bit ugly using the big if and I created a second one using a map, to make it shorter and easier, I would think.

const brackets = {
  '(': ')',
  '{': '}',
  '[': ']',
}
const openingBrackets = Object.keys(brackets)
const closingBrackets = Object.values(brackets)

function checkBrackets (string) {
  const bracketsStack = []
  charIterator = string[Symbol.iterator]()
  let char
  while (char = charIterator.next().value) {
    if (openingBrackets.includes(char)) {
      bracketsStack.push(char)
    } else if (closingBrackets.includes(char)) {
      if (brackets[bracketsStack.pop()] !== char) {
        return false
      }
    } 
  }
  return true
} 


console.log(checkBrackets('()()')) // true
console.log(checkBrackets('([])')) // true
console.log(checkBrackets('([)]')) // false
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What do you think? Is there a more elegant version to do it?

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