Validate the order of parenthese in an expression
katongole Isaac
Posted on January 3, 2023
I wrote this logic when i was developing a simple calculator. My aim was to validate the order of parenthese in an expression.
Please you can go through it and help me identify some bugs in it
Here is the implementation.
const brackets = ["(", ")"];
/**
* validates the order of parenthese in a string
* @param {*} type string
* @returns boolean otherwise `throws an error`
*/
const checkSyntax = (str) => {
if (typeof str !== "string") return false;
//check if both of the brackets dont exists
if (str.indexOf(brackets[0]) === -1 && str.indexOf(brackets[1]) === -1)
return true;
// check if one of the brackets exists.
if (
(str.indexOf(brackets[0]) === -1 && str.indexOf(brackets[1]) !== -1) ||
(str.indexOf(brackets[1]) === -1 && str.indexOf(brackets[0]) !== -1)
)
throw new Error(
`Syntax error: ${str[str.length - 1]} at index ${[str.length - 1]}`
);
//check if ) is at index 0 or ( is a last index throw otherwise skip
if (str[0] === brackets[1] || str[str.length - 1] === brackets[0])
throw new Error(`Error ==> index ${[str.length - 1]}`);
//Here we now check for validity
let stack = [];
for (let i = 0; i < str.length; i++) {
if (str[i] === brackets[0]) {
stack.push(str[i]);
continue;
}
if (str[i] === brackets[1]) {
//if we are trying to pop an empty stack []
if (!stack.length) throw new Error(`Error : not in order ${i}`);
stack.pop(str[i]);
//if the stack is not empty and i is on the last iteration then throw otherwise skip
if (stack.length !== 0 && i === str.length - 1)
throw new Error(
`Syntax Error : stack is not empty i.e stack length = [${stack.length}]`
);
continue;
}
}
return true;
};
console.log(checkSyntax("((()))()()()")); //true
💖 💪 🙅 🚩
katongole Isaac
Posted on January 3, 2023
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