Java 15

    public static String superReducedString(String s) {
        List<Character> chList = s.chars()
                               .mapToObj(c -> (char) c) 
                               .collect(Collectors.toList());
        int index = 0;
        while (index < chList.size() - 1) {
            if (chList.get(index).equals(chList.get(index + 1))) {
                chList.remove(index); 
                chList.remove(index);
                if (index > 0) index--;
            } else index++;
        }
        
        String result = chList.stream()
            .map(e -> e.toString()).collect(Collectors.joining());
        return result.length() > 0 ? result : "Empty String";
    }

C++ Trick is to use a stack, but can just use a string the same way. Moving from left to right, if the current char is the same as on the top of stack (or .back() of string) pop it (pop_back()), and do not push it to stack, effectivly deleting them. Else we push character to the stack (push_back()). If we use a string as a stack, we can just return this resulting string;

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

!/bin/python3

def superReducedString(s): # Write your code here stack = [] for char in s: if stack and stack[-1] == char: stack.pop() else: stack.append(char) reduced_string = ''.join(stack) return reduced_string if reduced_string else 'Empty String'

if name == 'main': fptr = open(os.environ['OUTPUT_PATH'], 'w')

s = input()

result = superReducedString(s)

fptr.write(result + '\n')

fptr.close()

My answer in Python

def superReducedString(s):
    print(s)
    pattern = r"(.)\1"
    
    while re.search(pattern, s):
        s = re.sub(pattern, "", s)
        print(s)
    
    if s:
        return s
    else:
        return "Empty String"