1. Abridged Problem Statement  
You start with an empty string and a cursor at the beginning. Given a sequence of key presses consisting of lowercase letters (`'a'–'z'`), `L` (move cursor left), and `R` (move cursor right), process each key:  
- Letter: insert it immediately after the cursor and move cursor right.  
- `L`: move cursor left if not already at the start.  
- `R`: move cursor right if not already at the end.  
After all operations, output the resulting string.

2. Detailed Editorial  
We need to process up to 1 000 000 operations in order, with cursor movements and insertions. Naïvely using a resizable array (e.g. `std::string`) and performing insertions/moves in the middle would be O(N²) in the worst case. Instead, we maintain two sequences that represent the text to the left and right of the cursor:

- `left`: characters to the left of the cursor, in order.
- `right`: characters to the right of the cursor, in order.

We start with both empty. As we read each operation `c`:

- If `c` is a lowercase letter, we append it to `left` (this is equivalent to inserting right after the cursor and moving cursor right).
- If `c == 'L'` and `left` is nonempty, we pop the last character from `left` and push it onto `right` (cursor moves left).
- If `c == 'R'` and `right` is nonempty, we pop the last character from `right` and push it onto `left` (cursor moves right).

All these operations on the end of a vector or deque are amortized O(1). At the end, the final text is `left` followed by the reverse of `right`.

Time complexity: O(N). Space complexity: O(N).

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << for pairs to ease debugging or printing (not used here).
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> for pairs.
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

// Overload >> for vectors.
template<typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for (auto &x: a) in >> x;
    return in;
}

// Overload << for vectors.
template<typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
    for (auto x: a) out << x << ' ';
    return out;
}

string s; // Input sequence of operations

// Read input into global string s
void read() {
    cin >> s;
}

// Solve using a doubly linked list and an iterator
void solve() {
    list<char> ans;            // Doubly linked list to store characters
    auto it = ans.begin();     // Iterator pointing to cursor position

    for (char c : s) {
        if (c == 'L') {
            // Move cursor left if possible
            if (it != ans.begin()) {
                --it;
            }
        } else if (c == 'R') {
            // Move cursor right if possible
            if (it != ans.end()) {
                ++it;
            }
        } else {
            // Insert character before iterator (which is after the cursor conceptually)
            ans.insert(it, c);
        }
    }

    // Output the entire list as the final text
    for (char c : ans) {
        cout << c;
    }
    cout << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
def main():
    data = sys.stdin.read().strip()
    # We'll treat 'left' and 'right' as stacks.
    left = []   # characters to the left of the cursor
    right = []  # characters to the right of the cursor

    for c in data:
        if c == 'L':
            if left:
                # Move one char from left to right (cursor moves left)
                right.append(left.pop())
        elif c == 'R':
            if right:
                # Move one char from right to left (cursor moves right)
                left.append(right.pop())
        else:
            # Letter insertion: push onto left (cursor moves right)
            left.append(c)

    # The final text is left + reversed(right)
    # right is in reverse order of actual text to its right
    sys.stdout.write(''.join(left) + ''.join(reversed(right)))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
Maintain two stacks (left/right of cursor). Letters go onto `left`; `L` pops from `left` to `right`; `R` pops from `right` to `left`. Final string is `left + reverse(right)`.