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 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 What do you think? Is there a more elegant version to do it?
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