Leetcode #2296: Design a Text Editor
In this guide, we solve Leetcode #2296 Design a Text Editor in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Design, Linked List, String, Doubly-Linked List, Simulation
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
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.
Python Solution
class TextEditor:
def __init__(self):
self.left = []
self.right = []
def addText(self, text: str) -> None:
self.left.extend(list(text))
def deleteText(self, k: int) -> int:
k = min(k, len(self.left))
for _ in range(k):
self.left.pop()
return k
def cursorLeft(self, k: int) -> str:
k = min(k, len(self.left))
for _ in range(k):
self.right.append(self.left.pop())
return ''.join(self.left[-10:])
def cursorRight(self, k: int) -> str:
k = min(k, len(self.right))
for _ in range(k):
self.left.append(self.right.pop())
return ''.join(self.left[-10:])
# Your TextEditor object will be instantiated and called as such:
# obj = TextEditor()
# obj.addText(text)
# param_2 = obj.deleteText(k)
# param_3 = obj.cursorLeft(k)
# param_4 = obj.cursorRight(k)
Complexity
The time complexity is . The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.