JavaScript Solution:

function superReducedString(s) {
   const splittedString = s.split('');
   for (var i = 0; i <= splittedString.length - 1; i++) {
      if (splittedString[i] === splittedString[i + 1]) {
         splittedString.splice(i, 2);
         i = -1;
      };
   }

   return splittedString.length ? splittedString.join('') : 'Empty String';
}

public static String superReducedString(String s) {
// Write your code here
    Stack<Character> stack = new Stack<>();
    for (int i = 0; i < s.length(); i++) {
        Character ch = s.charAt(i);
        if (!stack.isEmpty() && ch == stack.peek()) {
            stack.pop();
        } else {
            stack.push(ch);
        }
    }

    /* Print final result */

  StringBuilder result = new StringBuilder(); 

    while(!stack.isEmpty()){
        result.insert(0,stack.pop());
        System.out.println(result);
    }

    if(result.length()==0){
        return "Empty String";
    }
    return result.toString(); 

}

**Java Solution **

• public static String superReducedString(String s) {
• String str = "";
• int a ;
• for(int i = 0;i
• if(s.length()>1 && s.charAt(i) == s.charAt(i+1)){
• s = s.substring(0,i) + s.substring(i+2);
• a = -i;
• } else{
• a = 1;
• }
• }
• str = (s.length() == 0) ? "Empty String" : s;
• return str;
• }

Python Solution

def superReducedString(s):
    if len(s) < 2: return s
    
    i,n = 0,len(s)
    
    while i < n - 1:
        if s[i] == s[i+1]:
            s = s[:i] + s[i+2:]
            i,n = 0,len(s)
            continue
        
        i += 1
    
    return "Empty String" if n == 0 else s

Python solution

def superReducedString(s):
    i = 0
    s = list(s)
    while i < len(s)-1:
        if s[i]== s[i+1]:
            del s[i]
            del s[i]
            if i != 0: i -= 1 # to cover baab situation
        else:
             i += 1
             
    
    return (''.join(s)) if len(s) != 0 else "Empty String"