The C# code solution:

public static string superReducedString(string s) { List reduced = new List();

    foreach(char c in s)
    {
        if(reduced.Count>0 && reduced[reduced.Count -1] ==c)
        {
            reduced.RemoveAt(reduced.Count-1);
        }else{
            reduced.Add(c);
        }
    }

    if(reduced.Count==0) return "Empty String";
    else return new string(reduced.ToArray());

}

}

My Java 8 Solution

public static String superReducedString(String s) {
        StringBuilder reducedString = new StringBuilder();
        
        for (char c : s.toCharArray()) {
            int length = reducedString.length();
            
            if (length > 0 && reducedString.charAt(length - 1) == c) {
                reducedString.deleteCharAt(length - 1);
            } else {
                reducedString.append(c);
            }
        }
        
        return reducedString.length() == 0 ? "Empty String" : reducedString.toString();
    }

public static String superReducedString(String str) {
    String reducedString = _superReducedString(str);
    while(hasAdjacent(reducedString) && !reducedString.equals("Empty String")) {
        reducedString = _superReducedString(reducedString); 
    }
    return reducedString;
}

private static String _superReducedString(String str) {
    if(str == null || str.trim().equals("")) return "Empty String";
    if(str.length() == 1) return str;
    if(!hasAdjacent(str)) return str;

    if(str.substring(0, 1).equals(str.substring(1, 2))) {
        return _superReducedString(str.substring(2));
    }
    String s = _superReducedString(str.substring(1));
    return str.substring(0, 1) + (s.equals("Empty String") ? "" : s);       
}

static boolean hasAdjacent(String str) {
    for(int i=0; i<str.length()-1; i++) {
        if(str.charAt(i) == str.charAt(i+1)) return true;
    }
    return false;
}

here is a python solution with O(n) time and O(n) space in worse case; however, if the stack ends up being empty the space will be O(1)

def superReducedString(s):
    stack = []
    
    for c in s:
        if not stack or stack[-1] != c:
            stack.append(c)
        else:
            stack.pop()
    
    if not stack:
        return "Empty String"
        
    return "".join(stack)

Here is problem solution in Python, Java, C++, C and javascript - https://programmingoneonone.com/hackerrank-super-reduced-string-problem-solution.html