Design a Text Editor - Practice Coding Problems

Design a Text Editor - Problem

Linked List String Stack Design Simulation Doubly-Linked List Hard

Design a text editor with a cursor that can perform the following operations:

Add text to where the cursor is positioned
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 always remain within the actual text boundaries: 0 ≤ cursor.position ≤ currentText.length.

Implement the TextEditor class with these methods:

TextEditor() - Initializes the object with empty text
void addText(string text) - Appends text at cursor position, cursor moves to right of added text
int deleteText(int k) - Deletes k characters to the left of cursor, returns actual number of characters deleted
string cursorLeft(int k) - Moves cursor left k times, returns last min(10, len) characters to the left of cursor
string cursorRight(int k) - Moves cursor right k times, returns last min(10, len) characters to the left of cursor

Input & Output

Example 1 — Basic Operations

$ Input: operations = [["TextEditor"],["addText","leetcode"],["deleteText",4],["addText","practice"],["cursorRight",3],["cursorLeft",8]]

› Output: [null,null,4,null,"etpractice","leet"]

💡 Note: TextEditor textEditor = new TextEditor(); // Initially text="", cursor=0 textEditor.addText("leetcode"); // text="leetcode", cursor=8 textEditor.deleteText(4); // return 4, text="leet", cursor=4 textEditor.addText("practice"); // text="leetpractice", cursor=12 textEditor.cursorRight(3); // return "etpractice" (cursor at end, last 10 chars) textEditor.cursorLeft(8); // return "leet" (cursor at pos 4, 4 chars to left)

Example 2 — Cursor Movements

$ Input: operations = [["TextEditor"],["addText","hello"],["cursorLeft",3],["addText","world"],["cursorRight",2]]

› Output: [null,null,"he",null,"heworldl"]

💡 Note: Start empty, add "hello" (cursor at end), move left 3 positions (cursor after 'e'), add "world", move right 2 positions. Final text: "heworldllo"

Example 3 — Edge Case Boundaries

$ Input: operations = [["TextEditor"],["cursorLeft",5],["addText","a"],["deleteText",10],["cursorRight",5]]

› Output: [null,"",null,1,""]

💡 Note: Try to move left from empty (returns ""), add 'a', try to delete 10 chars (only deletes 1), try to move right from empty (returns "")

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

Visualization

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Understanding the Visualization

Input Operations

Sequence of text editor commands: add, delete, cursor moves

Process Commands

Execute each operation while maintaining cursor position

Output Results

Return results for methods that have return values

Key Takeaway

🎯 Key Insight: Two stacks perfectly simulate a text cursor - left stack holds text before cursor, right stack holds text after cursor, enabling O(1) operations

Asked in

G Google 25 M Microsoft 20 A Amazon 18 F Meta 15

Design a text editor using two stacks to efficiently handle cursor operations. The key insight is representing the cursor as the boundary between left and right stacks. Best approach achieves O(1) time for all operations. Time: O(1) per operation, Space: O(n) for text storage.

Common Approaches

Approach	Time	Space	Notes
✓ Single String with Rebuilding	O(n)	O(n)	Store text as single string, rebuild for every cursor operation
Two Stacks Approach	O(1)	O(n)	Use two stacks to represent text before and after cursor

Single String with Rebuilding — Algorithm Steps

Store text as single string with cursor index
For addText: slice string and insert at cursor
For deleteText: remove characters before cursor
For cursor moves: adjust index and extract substring

Visualization

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Step-by-Step Walkthrough

Initial State

Empty string with cursor at position 0

Add Text

Insert text at cursor position using string slicing

Move Cursor

Update cursor index and return substring

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_TEXT_LEN 100000

typedef struct {
    char text[MAX_TEXT_LEN];
    int cursor;
    int length;
} TextEditor;

TextEditor* textEditorCreate() {
    TextEditor* editor = (TextEditor*)malloc(sizeof(TextEditor));
    editor->text[0] = '\0';
    editor->cursor = 0;
    editor->length = 0;
    return editor;
}

void textEditorAddText(TextEditor* editor, char* text) {
    int textLen = strlen(text);
    memmove(editor->text + editor->cursor + textLen, editor->text + editor->cursor, editor->length - editor->cursor + 1);
    memcpy(editor->text + editor->cursor, text, textLen);
    editor->cursor += textLen;
    editor->length += textLen;
}

int textEditorDeleteText(TextEditor* editor, int k) {
    int deleted = (k < editor->cursor) ? k : editor->cursor;
    memmove(editor->text + editor->cursor - deleted, editor->text + editor->cursor, editor->length - editor->cursor + 1);
    editor->cursor -= deleted;
    editor->length -= deleted;
    return deleted;
}

char* textEditorCursorLeft(TextEditor* editor, int k) {
    editor->cursor = (editor->cursor > k) ? editor->cursor - k : 0;
    static char result[11];
    int start = (editor->cursor > 10) ? editor->cursor - 10 : 0;
    strncpy(result, editor->text + start, editor->cursor - start);
    result[editor->cursor - start] = '\0';
    return result;
}

char* textEditorCursorRight(TextEditor* editor, int k) {
    int newCursor = editor->cursor + k;
    editor->cursor = (newCursor < editor->length) ? newCursor : editor->length;
    static char result[11];
    int start = (editor->cursor > 10) ? editor->cursor - 10 : 0;
    strncpy(result, editor->text + start, editor->cursor - start);
    result[editor->cursor - start] = '\0';
    return result;
}

int main() {
    // Simplified implementation for demonstration
    TextEditor* editor = textEditorCreate();
    
    // Parse operations and execute
    // This would require proper JSON parsing in real implementation
    
    free(editor);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Each operation may require O(n) string slicing and concatenation

✓ Linear Growth

Space Complexity

O(n)

Stores the text string of length n

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single String with Rebuilding — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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