How to implement count missing brackets?

Question

For example: I have a string only contains brackets '(' and ')'. How to count missing brackets?

My code:

var str = '(())';
var open = 0, close = 0, count = 0;
for (var i = 0; i < str.length; i++) {
  if (str[i] == '(') {
    open++;
    count = Math.abs(open - close);
  } else {
    close++;
    count = Math.abs(open - close);
  }
}
console.log(count);

Input: '(())' Output: 0

Input: '(()' Output: 1

Input: '))((' Output: 4

Input: '(()(' Output: 2

Input: '(()()))(())(())' Output: 1

Answer 1

You could replace perfect brackets and count the rest.

function missing(string) {
    var l;

    do {
        l = string.length;
        string = string.replace(/\(\)/g, '');
    } while (l !== string.length)
 
    return string.length;
}

console.log(missing('(())')); // 0
console.log(missing('(()')); // 1
console.log(missing('))((')); // 4
console.log(missing('(()(')); // 2
console.log(missing('(()()))(())(())')); // 3

Answer 2

Below code output missing parenthesis count

var missingParenthesisCnt = function(s) {
       var stack = 0, combo=[];
      for (var i = 0; i < s.length; i++) {
          var item = s[i];
          if (s[i]=="(") {
            if(stack<0){
                combo.push(Math.abs(stack))
                stack=0
            }
            stack+=1
          } 
          else if(s[i]==")"){
            stack-=1
          }
        }
      return stack + combo.reduce((a,b)=>a+b, 0);
    }
    console.log("Total Missing Parenthesis Count:", missingParenthesisCnt(")))((()"))