chore: add LeetCode daily solution

Author: TechnoBlogger14o3

Medium/2025-09-19-3484-Design-Spreadsheet/explanation.md

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+# Design Spreadsheet (Medium)
+
+**Problem ID:** 3484  
+**Date:** 2025-09-19  
+**Link:** https://leetcode.com/problems/design-spreadsheet/
+
+## Approach
+
+To solve the "Design Spreadsheet" problem, we can adopt an object-oriented approach by implementing a `Spreadsheet` class that encapsulates the required functionalities. Here's a concise breakdown of the approach:
+
+### Main Idea:
+The goal is to create a spreadsheet that supports setting and resetting cell values, along with evaluating simple arithmetic formulas involving cell references and integers.
+
+### Data Structures:
+1. **2D Array (or List of Lists)**: To represent the spreadsheet, we can use a 2D list where each element corresponds to a cell in the spreadsheet. The rows will be indexed from 0 to `rows - 1`, and the columns will be represented by indices corresponding to letters 'A' to 'Z' (0 to 25).
+   
+2. **Dictionary (or HashMap)**: To facilitate quick access to cell values, we can maintain a dictionary that maps cell references (like "A1", "B2") to their respective values. This will allow us to handle cases where cells have not been explicitly set, defaulting to 0.
+
+### Methods:
+1. **Constructor (`Spreadsheet(int rows)`)**: Initializes the spreadsheet with the specified number of rows and sets all cell values to 0.
+
+2. **`setCell(String cell, int value)`**: Converts the cell reference into its respective row and column indices, updates the value in the 2D list, and also updates the dictionary.
+
+3. **`resetCell(String cell)`**: Similar to `setCell`, but sets the specified cell's value back to 0 in both the 2D list and the dictionary.
+
+4. **`getValue(String formula)`**: This method parses the formula string to extract the components (X and Y). It checks if they are cell references or integers, retrieves their values (defaulting to 0 for unset cells), and computes the sum. The parsing can be done using string manipulation techniques.
+
+### Complexity:
+- **Time Complexity**:
+  - `setCell`: O(1) for direct updates.
+  - `resetCell`: O(1) for direct resets.
+  - `getValue`: O(1) for retrieving values, but parsing the formula may take O(n) where n is the length of the formula string. In the worst case, this could be O(1) since the formula is always in the format "=X+Y" with a fixed length.
+  
+- **Space Complexity**: O(rows * 26) for the 2D array and O(1) for the dictionary, as it only stores values for cells that have been set.
+
+### Conclusion:
+This approach efficiently manages the spreadsheet's state and allows for quick updates and evaluations of cell values, adhering to the constraints provided. The use of a 2D array and a dictionary ensures that operations are performed in constant time for most cases, making the solution both effective and scalable.

Medium/2025-09-19-3484-Design-Spreadsheet/solution.java

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+import java.util.HashMap;
+import java.util.Map;
+
+class Spreadsheet {
+    private int rows;
+    private int[][] cells;
+    private Map<String, Integer> cellMap;
+
+    public Spreadsheet(int rows) {
+        this.rows = rows;
+        this.cells = new int[rows][26];
+        this.cellMap = new HashMap<>();
+    }
+
+    public void setCell(String cell, int value) {
+        int row = Integer.parseInt(cell.substring(1)) - 1;
+        int col = cell.charAt(0) - 'A';
+        cells[row][col] = value;
+        cellMap.put(cell, value);
+    }
+
+    public void resetCell(String cell) {
+        int row = Integer.parseInt(cell.substring(1)) - 1;
+        int col = cell.charAt(0) - 'A';
+        cells[row][col] = 0;
+        cellMap.put(cell, 0);
+    }
+
+    public int getValue(String formula) {
+        if (formula.charAt(0) == '=') {
+            String[] parts = formula.substring(1).split("\\+");
+            int sum = 0;
+            for (String part : parts) {
+                if (Character.isDigit(part.charAt(0))) {
+                    sum += Integer.parseInt(part);
+                } else {
+                    sum += cellMap.getOrDefault(part, 0);
+                }
+            }
+            return sum;
+        }
+        return 0;
+    }
+}

Medium/2025-09-19-3484-Design-Spreadsheet/solution.js

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+class Spreadsheet {
+    constructor(rows) {
+        this.rows = rows;
+        this.cells = new Map();
+    }
+
+    setCell(cell, value) {
+        this.cells.set(cell, value);
+    }
+
+    resetCell(cell) {
+        this.cells.delete(cell);
+    }
+
+    getValue(formula) {
+        if (formula[0] === '=') {
+            const parts = formula.slice(1).split('+');
+            let total = 0;
+            for (const part of parts) {
+                const trimmedPart = part.trim();
+                const num = parseInt(trimmedPart);
+                if (!isNaN(num)) {
+                    total += num;
+                } else {
+                    total += this.cells.get(trimmedPart) || 0;
+                }
+            }
+            return total;
+        }
+        return 0;
+    }
+}

Medium/2025-09-19-3484-Design-Spreadsheet/solution.py

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+class Spreadsheet:
+    def __init__(self, rows: int):
+        self.rows = rows
+        self.cells = {}
+    
+    def setCell(self, cell: str, value: int) -> None:
+        self.cells[cell] = value
+    
+    def resetCell(self, cell: str) -> None:
+        if cell in self.cells:
+            del self.cells[cell]
+    
+    def getValue(self, formula: str) -> int:
+        if formula[0] == '=':
+            formula = formula[1:]
+        parts = formula.split('+')
+        total = 0
+        for part in parts:
+            part = part.strip()
+            if part.isdigit():
+                total += int(part)
+            else:
+                total += self.cells.get(part, 0)
+        return total

README.md

@@ -151,3 +151,4 @@ Each problem includes:
 - 2025-09-15 — [Maximum Number of Words You Can Type](https://leetcode.com/problems/maximum-number-of-words-you-can-type/) (Easy) → `Easy/2025-09-15-1935-Maximum-Number-of-Words-You-Can-Type`
 - 2025-09-17 — [Design a Food Rating System](https://leetcode.com/problems/design-a-food-rating-system/) (Medium) → `Medium/2025-09-17-2353-Design-a-Food-Rating-System`
 - 2025-09-18 — [Design Task Manager](https://leetcode.com/problems/design-task-manager/) (Medium) → `Medium/2025-09-18-3408-Design-Task-Manager`
+- 2025-09-19 — [Design Spreadsheet](https://leetcode.com/problems/design-spreadsheet/) (Medium) → `Medium/2025-09-19-3484-Design-Spreadsheet`