leetcode

Design Spreadsheet - Link

Question Description

Implement a simple spreadsheet where:

• You can set a cell with a numeric value
• You can reset a cell (making its value 0)
• You can evaluate a formula string of the form “A1+B2+3”, where tokens may be integers or cell references

Methods:

• Spreadsheet(int rows) → initializes spreadsheet
• void setCell(String cell, int value) → sets a cell value
• void resetCell(String cell) → resets cell to 0
• int getValue(String formula) → evaluates a formula string

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Constraints

• 1 <= rows <= 1000
• Cell references are in format “A1”, “B2”, etc. (A-Z for columns, 1-1000 for rows)
• Formula format: “=A1+B2+3” (starts with =, terms separated by +)
• 1 <= values <= 10^6

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Approach

Use a HashMap to store cell values with cell references as keys.

Operations:

• setCell: Store value in HashMap with cell reference as key
• resetCell: Set cell value to 0 in HashMap
• getValue: Parse formula string, evaluate each term, sum results

Formula parsing:

• Remove ‘=’ prefix
• Split by ‘+’ to get individual terms
• For each term, check if it starts with letter (cell reference) or digit (number)
• Cell references: get value from HashMap (default 0 if not set)
• Numbers: parse as integer

Why this approach works:

• HashMap provides O(1) average lookup for cell values
• Simple string parsing handles the formula format efficiently
• Default value of 0 for unset cells matches spreadsheet behavior

Alternative approaches considered:

• Could use 2D array for fixed rows, but HashMap is more flexible
• Could use more complex expression parser, but simple split is sufficient

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Dry Run

Example Operations:

Spreadsheet obj = new Spreadsheet(3);
obj.setCell("A1", 5);
obj.setCell("B1", 10);
obj.getValue("=A1+B1+3"); // returns 18
obj.resetCell("A1");
obj.getValue("=A1+B1+3"); // returns 13

Step-by-step execution:

• Step 1: Initialize empty HashMap
• Step 2: setCell(“A1”, 5) → HashMap: {“A1”: 5}
• Step 3: setCell(“B1”, 10) → HashMap: {“A1”: 5, “B1”: 10}
• Step 4: getValue(“=A1+B1+3”):
  • Parse: [“A1”, “B1”, “3”]
  • A1 = 5, B1 = 10, 3 = 3
  • Sum = 18
• Step 5: resetCell(“A1”) → HashMap: {“A1”: 0, “B1”: 10}
• Step 6: getValue(“=A1+B1+3”):
  • Parse: [“A1”, “B1”, “3”]
  • A1 = 0, B1 = 10, 3 = 3
  • Sum = 13

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Solution

import java.util.*;

class Spreadsheet {

    private Map<String, Integer> cellMap = new HashMap<>();
    
    public Spreadsheet(int rows) {
        // rows parameter can be used for validation if needed
    }
    
    public void setCell(String cell, int value) {
        cellMap.put(cell, value);
    }
    
    public void resetCell(String cell) {
        cellMap.put(cell, 0);
    }
    
    public int getValue(String formula) {
        int sum = 0;
        // skip '=' at start
        for (String s : formula.substring(1).split("\\+")) {
            sum += mapToValue(s.trim());
        }
        return sum;
    }

    private int mapToValue(String s) {
        return Character.isLetter(s.charAt(0)) ? cellMap.getOrDefault(s, 0) : Integer.parseInt(s);
    }
}

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Time and Space Complexity

• Time Complexity:
  • setCell, resetCell: O(1) average (hash map put/get)
  • getValue: O(k) where k = number of terms in the formula
• Space Complexity: O(n) for storing n cells in the map

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