replaceParenthesis

# You are given input string containing alphabetical chaterters and parenthesis. Convert the input string into following format.

# "()" => "00"
# ")(" => "-11"
# "((ab)" => "10ab0"
# "(a(b)c)" => "0a0b0c0"
# "()a((((" => "00a1111"
# "(a))))b(((" => "0a0-1-1-1b111"


# (a(b)c)
# Stack: [0a0b0c0]
 

# Replace both opening and closing braces with 0 if parenthesis are balanced
# If parenthesis are not balance, replace '(' -> 1 and ')' -> -1.





let str="((ab)"

function replaceParenthesis(str){
  let stack=[]
  for(let i=0; i<str.length; i++){
   let temp='',top// reinitialize 
    if(str[i]==')'){
       temp=')'
      while(top!=='(' && stack.length>0){
        top=stack.pop()
        temp=top+temp//opposite order append 
      }
      
     // just for the balanced replacement
      if(temp.charAt(temp.length-1)==')' && temp.charAt(0)=='('){
        temp=temp.replace('(','0')
        temp=temp.replace(')','0')
      }
      for(let j=0; j<temp.length; j++){
        stack.push(temp[j])//push all the pop element into the stack 
      }
    } 
    else stack.push(str[i])
  }

  //now assign imbalance mapping, replace '(' -> 1 and ')' -> -1
  for(let k=0;k<stack.length;k++){
    if(stack[k]=='(') stack[k]='1'
    else if(stack[k]==')')stack[k]='-1'
  }
  stack=stack.join('')
  console.log('Finalstack',stack)
  }

replaceParenthesis(str)