LeetCode Problem Workspace

Design Spreadsheet

Design a spreadsheet that supports setting, retrieving, and resetting values, with the ability to handle formulas referencing other cells.

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Practice Focus

Medium · Array scanning plus hash lookup

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Answer-first summary

Design a spreadsheet that supports setting, retrieving, and resetting values, with the ability to handle formulas referencing other cells.

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This problem asks you to design a spreadsheet that can handle setting values, resetting cells, and performing basic arithmetic formulas. The challenge lies in handling the interdependencies between cells through efficient lookups and updates, which can be achieved using a hashmap to represent cell references and their values.

Problem Statement

Design a spreadsheet with 26 columns labeled 'A' to 'Z' and a given number of rows. Each cell can store an integer value between 0 and 105. Your task is to implement a Spreadsheet class capable of handling operations like setting cell values, retrieving them, and resetting them to zero. Additionally, cells may contain formulas of the form '=X+Y', where X and Y are either valid cell references or integers.

If a cell references another cell that has not been explicitly set, its value is considered 0. You need to handle the formulas dynamically, ensuring correct calculations when referencing other cells. The challenge includes managing the grid structure efficiently and ensuring that updates to one cell reflect correctly when other cells are referenced.

Examples

Example 1

Input: ["Spreadsheet", "getValue", "setCell", "getValue", "setCell", "getValue", "resetCell", "getValue"] [[3], ["=5+7"], ["A1", 10], ["=A1+6"], ["B2", 15], ["=A1+B2"], ["A1"], ["=A1+B2"]]

Output: [null, 12, null, 16, null, 25, null, 15] Explanation

Example details omitted.

Constraints

  • 1 <= rows <= 103
  • 0 <= value <= 105
  • The formula is always in the format "=X+Y", where X and Y are either valid cell references or non-negative integers with values less than or equal to 105.
  • Each cell reference consists of a capital letter from 'A' to 'Z' followed by a row number between 1 and rows.
  • At most 104 calls will be made in total to setCell, resetCell, and getValue.

Solution Approach

Using HashMap for Cell References

You can represent the cells in the spreadsheet using a hashmap where the key is the cell reference (e.g., 'A1') and the value is the integer stored in that cell. This approach allows constant time lookups and updates, making it efficient to handle the operations setCell, getValue, and resetCell.

Handling Formulas Dynamically

For cells with formulas, parse and evaluate the formula by retrieving the referenced cell values and performing the arithmetic. If the formula references other cells, ensure the dependent cells are computed first. This ensures correct results for cells that depend on others.

Optimizing with Memoization

To avoid recalculating cell values that depend on formulas multiple times, use memoization to store the evaluated values. When a cell is updated or reset, ensure all dependent cells are re-evaluated correctly to maintain consistency in the spreadsheet.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend on how the formulas are evaluated. If we use a simple hashmap, each setCell, resetCell, and getValue operation takes constant time. However, evaluating formulas might require scanning dependent cells, which could increase the time complexity. Memoization can help improve performance by reducing redundant calculations.

What Interviewers Usually Probe

  • Can the candidate efficiently use hashmaps for quick cell reference lookup and updates?
  • Does the candidate handle dynamic formula evaluation and interdependencies correctly?
  • Is the candidate aware of optimization techniques, like memoization, to avoid redundant calculations?

Common Pitfalls or Variants

Common pitfalls

  • Forgetting to handle cell references that are not explicitly set, leading to incorrect results when retrieving values.
  • Not handling the resetCell operation correctly, which might cause inconsistent results when a cell is reset but still referenced in formulas.
  • Improperly evaluating formulas, especially when dependent cells are not updated in the correct order, leading to incorrect values.

Follow-up variants

  • Allowing formulas to include subtraction or multiplication in addition to addition.
  • Supporting larger grids, such as having more than 26 columns or more rows.
  • Handling circular dependencies in formulas and preventing infinite loops in evaluations.

FAQ

How does the spreadsheet handle formulas referencing unset cells?

If a formula references a cell that has not been explicitly set, its value is considered 0 by default.

What is the time complexity of evaluating a formula?

The time complexity depends on how many cells the formula references. A simple reference would be O(1), but more complex formulas could require O(n) time if there are multiple dependent cells.

Can the spreadsheet handle circular dependencies in formulas?

No, the current problem doesn't specify handling circular dependencies, so it assumes formulas do not cause such loops. You can handle this with additional logic if needed.

How should I implement the resetCell operation?

The resetCell operation should set the referenced cell’s value back to zero. Ensure that any formulas relying on this cell are updated or recalculated accordingly.

What is the most efficient way to store cell values?

Using a hashmap where the key is the cell reference (e.g., 'A1') and the value is the integer stored in the cell is the most efficient way to handle quick lookups and updates.

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Solution

Solution 1: Hash Table

We use a hash table $\textit{d}$ to record the values of all cells, where the key is the cell reference and the value is the cell's value.

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class Spreadsheet:

    def __init__(self, rows: int):
        self.d = {}

    def setCell(self, cell: str, value: int) -> None:
        self.d[cell] = value

    def resetCell(self, cell: str) -> None:
        self.d.pop(cell, None)

    def getValue(self, formula: str) -> int:
        ans = 0
        for cell in formula[1:].split("+"):
            ans += int(cell) if cell[0].isdigit() else self.d.get(cell, 0)
        return ans


# Your Spreadsheet object will be instantiated and called as such:
# obj = Spreadsheet(rows)
# obj.setCell(cell,value)
# obj.resetCell(cell)
# param_3 = obj.getValue(formula)
Design Spreadsheet Solution: Array scanning plus hash lookup | LeetCode #3484 Medium