How do you solve Design Spreadsheet on LeetCode?

Design Spreadsheet (LeetCode #3484) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hashmap optimizes cell access.

What is the time complexity of Design Spreadsheet?

The optimal solution for Design Spreadsheet runs in O(n) time and uses O(n) space. The hashmap allows for constant time access to cell values, making operations efficient.

What is the brute force solution for Design Spreadsheet?

The brute force approach for Design Spreadsheet is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach directly manipulates a 2D array to represent the spreadsheet. Each operation is performed on the array, which can be inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Spreadsheet?

Design Spreadsheet uses the following concepts: Array, Hash Table, String, Design, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design Spreadsheet?

Clarify the problem requirements. Discuss edge cases. Explain your thought process clearly.

What are common mistakes when solving Design Spreadsheet?

Forgetting to handle missing cells in getValue. Incorrectly parsing cell references.

#3484

Design Spreadsheet

Medium

ArrayHash TableStringDesignMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a hashmap allows for efficient cell access and updates, reducing the need for a full grid representation.

⚙️

Algorithm

3 steps

1Step 1: Initialize a hashmap to store cell values, where keys are cell references.
2Step 2: For setCell, directly update the hashmap with the cell reference and value.
3Step 3: For getValue, parse the formula and retrieve values from the hashmap, handling missing cells by returning 0.

solution.py10 lines

1class Spreadsheet:
2    def __init__(self, rows):
3        self.cells = {}
4    def setCell(self, cell, value):
5        self.cells[cell] = value
6    def resetCell(self, cell):
7        self.setCell(cell, 0)
8    def getValue(self, formula):
9        x, y = formula[1:].split('+')
10        return self.cells.get(x, 0) + self.cells.get(y, 0)

ℹ

Complexity note: The hashmap allows for constant time access to cell values, making operations efficient.

1Using a hashmap optimizes cell access.
2Resetting cells is straightforward with direct updates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.