Introduction

In this blog post, we will explore how to solve the LeetCode problem "2296" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Design a text editor with a cursor that can do the following: Add text to where the cursor is. Delete text from where the cursor is (simulating the backspace key). Move the cursor either left or right. When deleting text, only characters to the left of the cursor will be deleted. The cursor will also remain within the actual text and cannot be moved beyond it. More formally, we have that 0 <= cursor.position <= currentText.length always holds. Implement the TextEditor class: TextEditor() Initializes the object with empty text. void addText(string text) Appends text to where the cursor is. The cursor ends to the right of text. int deleteText(int k) Deletes k characters to the left of the cursor. Returns the number of characters actually deleted. string cursorLeft(int k) Moves the cursor to the left k times. Returns the last min(10, len) characters to the left of the cursor, where len is the number of characters to the left of the cursor. string cursorRight(int k) Moves the cursor to the right k times. Returns the last min(10, len) characters to the left of the cursor, where len is the number of characters to the left of the cursor. Example 1: Input ["TextEditor", "addText", "deleteText", "addText", "cursorRight", "cursorLeft", "deleteText", "cursorLeft", "cursorRight"] [[], ["leetcode"], [4], ["practice"], [3], [8], [10], [2], [6]] Output [null, null, 4, null, "etpractice", "leet", 4, "", "practi"] Explanation TextEditor textEditor = new TextEditor(); // The current text is "|". (The '|' character represents the cursor) textEditor.addText("leetcode"); // The current text is "leetcode|". textEditor.deleteText(4); // return 4 // The current text is "leet|". // 4 characters were deleted. textEditor.addText("practice"); // The current text is "leetpractice|". textEditor.cursorRight(3); // return "etpractice" // The current text is "leetpractice|". // The cursor cannot be moved beyond the actual text and thus did not move. // "etpractice" is the last 10 characters to the left of the cursor. textEditor.cursorLeft(8); // return "leet" // The current text is "leet|practice". // "leet" is the last min(10, 4) = 4 characters to the left of the cursor. textEditor.deleteText(10); // return 4 // The current text is "|practice". // Only 4 characters were deleted. textEditor.cursorLeft(2); // return "" // The current text is "|practice". // The cursor cannot be moved beyond the actual text and thus did not move. // "" is the last min(10, 0) = 0 characters to the left of the cursor. textEditor.cursorRight(6); // return "practi" // The current text is "practi|ce". // "practi" is the last min(10, 6) = 6 characters to the left of the cursor. Constraints: 1 <= text.length, k <= 40 text consists of lowercase English letters. At most 2 * 104 calls in total will be made to addText, deleteText, cursorLeft and cursorRight. Follow-up: Could you find a solution with time complexity of O(k) per call?

Explanation

Here's a solution that addresses the prompt's requirements, focusing on efficiency:

• Two-Stack Approach: Maintain two stacks, left and right. The left stack stores characters to the left of the cursor, and the right stack stores characters to the right of the cursor. This allows O(1) insertion and deletion at the cursor position.
• Cursor Movement: Cursor movement is simulated by transferring elements between the left and right stacks. This keeps the cursor position implicit based on the stack boundaries.

• Result Retrieval: For returning the last 10 characters, we temporarily move those characters to a helper list to generate the string and then put them back on the stack, ensuring minimal disruption to the main data structure.

• Runtime Complexity: O(k) for deleteText, cursorLeft, and cursorRight, where k is the number of characters to delete or move. O(m) for addText, where m is the length of the added text. Storage Complexity: O(N), where N is the length of the text in the editor.

Code

    class TextEditor:

    def __init__(self):
        self.left = []
        self.right = []

    def addText(self, text: str) -> None:
        for char in text:
            self.left.append(char)

    def deleteText(self, k: int) -> int:
        deleted_count = 0
        while self.left and deleted_count < k:
            self.left.pop()
            deleted_count += 1
        return deleted_count

    def cursorLeft(self, k: int) -> str:
        moved_count = 0
        while self.left and moved_count < k:
            self.right.append(self.left.pop())
            moved_count += 1

        return "".join(self.left[-10:])

    def cursorRight(self, k: int) -> str:
        moved_count = 0
        while self.right and moved_count < k:
            self.left.append(self.right.pop())
            moved_count += 1
        return "".join(self.left[-10:])