Brackets

Given an expression string exp, write a program to examine whether the pairs and the orders of “{“, “}”, “(“, “)”, “[“, “]” are correct in exp.

Example:
console.log(Brackets(“()[]{}”)) 1
console.log(Brackets(“([{}])”)) 1
console.log(Brackets(“()]]”)) 0

function Brackets(input) {
  let stack = [];
  for(const c of input) {
    if (c === '{' || c === '[' || c === '(' ){
      stack.push(c);
    } else if (c ===  '}') {
      if(stack.length === 0 || stack.pop() !== '{') return 0;
    }else if (c ===  ']') {
      if(stack.length === 0 || stack.pop() !== '[') return 0;
    }else if (c ===  ')') {
      if(stack.length === 0 || stack.pop() !== '(') return 0;
}
    }
    return stack.length ? 0 : 1
  }

console.log(Brackets("()[]()[]{}"))
console.log(Brackets("([{}])"))
console.log(Brackets("()]]"))

We create a function called Brackets with parameter called input and we create an array called stack to hold all of the brackets;
We loop with for of loop where we create variable c of input and we check if c is ‘(‘ or c is ‘{‘ or c is ‘[’ opening parenthesis. And if we find we push into the array variable stack with JavaScript push method.
Then we create 3 if else statements for each of the individual parenthesis so we have else if we have closed parenthesis
1st else if (c === ‘}’) We check if the stack.length === 0 or stack.pop() pop() is another JavaScript method that take the last added in this case stack.pop() is not equal to the ‘{‘ we return 0 which means it faild
2nd else if (c === ‘]’) We check if the stack.length === 0 or stack.pop() pop() is another JavaScript method that take the last added in this case stack.pop() is not equal to the ‘[’ we return 0 which means it faild
3rd else if (c === ‘)’) We check if the stack.length === 0 or stack.pop() pop() is another JavaScript method that take the last added in this case stack.pop() is not equal to the ‘(‘ we return 0 which means it faild
At the end we return stack.length ? (true => there is something left) 0 : 1