!/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"

public static string superReducedString(string s)
{
    if(s.Length == 1) return s;        
    var i=1;
    while(i < s.Length){
        if(s.Substring(i-1, 1) == s.Substring(i, 1)){
            s = s.Remove(i-1, 2);
            if(i>1) i--;
        }
        else {
            i++;
        }
    }
    return s == "" ? "Empty String" : s;
}

Only three lines: Some practices in Hackerrank seem more difficult than easy and somes seem easier than intermediate.

def superReducedString(s): # Write your code here

for letter in s:
    s = s.replace(2*letter,"")

return s if len(s) != 0 else "Empty String"

Java O(n)

public static String superReducedString(String s) {
        Stack<Character> stack = new Stack<>();
        
        for (char c : s.toCharArray()) {
            if (!stack.isEmpty() && stack.peek() == c) {
                stack.pop();
            } else {
                stack.push(c);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for (char c : stack) {
            sb.append(c);
        }
        
        return sb.length() == 0 ? "Empty String" : sb.toString();
    }