My python solution

if name == 'main': text = [] edition = [] for _ in range(int(input())): op = input().split() if op[0] == '1': text.extend(op[1]) edition.append(len(op[1])) elif op[0] == '2': edition.append([]) for _ in range(int(op[1])): edition[-1].append(text.pop()) elif op[0] == '3': print(text[int(op[1])-1]) elif op[0] == '4': undo = edition.pop() if isinstance(undo, int): for _ in range(undo): text.pop() else: text.extend(undo[::-1])

I am at a loss on this one. Many of the test cases are failing, yet my output when I run locally matches the expected output perfectly. I get 'wrong answer'. If anyone has a clue what it could be let me know. My code:

private static final Scanner scanner = new Scanner(System.in);
private static Stack<String> state = new Stack<>();

public static void main(String[] args) throws IOException {
    state.push("");
    while (scanner.hasNext()) {
        String command = scanner.nextLine();
        int op = command.charAt(0) - '0';
        switch (op) {
            case 1:
                String toAppend = command.substring(2);
                state.push(append(state.peek(), toAppend));
                break;
            case 2:
                int charsToDelete = Integer.parseInt(command.substring(2));
                state.push(delete(state.peek(), charsToDelete));
                break;
            case 3:
                int charToPrint = Integer.parseInt(command.substring(2));
                System.out.println(state.peek().charAt(charToPrint - 1));
                break;
            case 4:
                state.pop();
                break;
        }

    }
}

private static String append(String text, String toAppend) {
    return text + toAppend;
}

private static String delete(String text, int charsToDelete) {
    return text.substring(0, text.length() - charsToDelete);
}

straightforward solution in python:

intial_input = input().split(" ")
n = int(intial_input[0])
if len(intial_input) > 1:
    s = str(intial_input[1])
else:
    s = ""
undo = []
for i in range(n):
    operation = input().split(" ")
    if operation[0] == '1':
        undo.append(s)
        s = s + (operation[1])
    elif operation[0] == '2':
        undo.append(s)
        x = len(s)
        s = s[0:x-int(operation[1])]
    elif operation[0] == '3':
        x = int(operation[1])-1
        print(s[x])
    elif operation[0] == '4':
        s = undo.pop()
  

This doesn't look like an intermediate level problem. Seems much easier that 'Lego Blocks' for example.

include

include

include

include

include

include

using namespace std;

class S { private: string my_string{};

public: 

void append (string str){
    my_string.append(str);
}

void deletechar (int k){
    my_string.erase(my_string.size() - k);
}

void print(int k){
    cout << my_string[k-1] <<endl;
}

//constructor
S (string str)
:my_string{str} {}

S(){}

S &operator= ( const S &lhs){
    this->my_string = lhs.my_string;
    return *this;
}

};

int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int Q; cin >> Q;

    int op;
    S my_string;

    stack<S> stk;

   while(Q--){
     cin >> op;
     string str;
     int k;
     switch(op) {
         case 1: 
         cin >> str;
         my_string.append(str);
         stk.push(my_string);
         break;

         case 2:
         cin >> k;
         my_string.deletechar(k);
         stk.push(my_string);
         break;

         case 3:
         cin >> k;
         my_string.print(k);
         break;

         case 4:
         if (!stk.empty()) {
                stk.pop();
                if (!stk.empty()) {
                    my_string = stk.top();
                } else {
                    my_string = S(); // Reset to default state if the stack is empty
                }
            }
            break;
         break;

         default:
         return 0;
     }  

   }
return 0;

}