title: Categorize Box According to Criteria slug: categorize-box-according-to-criteria difficulty: easy leetcode_id: 2525 leetcode_url: https://leetcode.com/problems/categorize-box-according-to-criteria/ categories: - math patterns: - greedy function_signature: "def categorize_box(length: int, width: int, height: int, mass: int) -> str:" test_cases: visible: - input: { length: 1000, width: 35, height: 700, mass: 300 } expected: "Heavy" - input: { length: 200, width: 50, height: 800, mass: 50 } expected: "Neither" hidden: - input: { length: 10000, width: 1, height: 1, mass: 1 } expected: "Bulky" - input: { length: 1000, width: 1000, height: 1000, mass: 100 } expected: "Both" - input: { length: 100, width: 100, height: 100, mass: 99 } expected: "Neither" - input: { length: 1, width: 10000, height: 1, mass: 200 } expected: "Both" - input: { length: 9999, width: 9999, height: 9999, mass: 50 } expected: "Bulky" description: | Given four integers `length`, `width`, `height`, and `mass`, representing the dimensions and mass of a box, respectively, return *a string representing the **category** of the box*. The box is `"Bulky"` if: - **Any** of the dimensions of the box is greater or equal to `10^4`. - Or, the **volume** of the box is greater or equal to `10^9`. If the mass of the box is greater or equal to `100`, it is `"Heavy"`. If the box is both `"Bulky"` and `"Heavy"`, then its category is `"Both"`. If the box is neither `"Bulky"` nor `"Heavy"`, then its category is `"Neither"`. If the box is `"Bulky"` but not `"Heavy"`, then its category is `"Bulky"`. If the box is `"Heavy"` but not `"Bulky"`, then its category is `"Heavy"`. **Note** that the volume of the box is the product of its length, width, and height. constraints: | - `1 <= length, width, height <= 10^5` - `1 <= mass <= 10^3` examples: - input: "length = 1000, width = 35, height = 700, mass = 300" output: '"Heavy"' explanation: "None of the dimensions is >= 10^4. Volume = 24,500,000 < 10^9, so not Bulky. But mass = 300 >= 100, so Heavy. Since not Bulky but Heavy, return \"Heavy\"." - input: "length = 200, width = 50, height = 800, mass = 50" output: '"Neither"' explanation: "None of the dimensions is >= 10^4. Volume = 8,000,000 < 10^9, so not Bulky. Mass = 50 < 100, so not Heavy. Since neither Bulky nor Heavy, return \"Neither\"." explanation: intuition: | Think of this problem like a sorting machine at a shipping warehouse. Packages arrive on a conveyor belt, and the machine needs to route them to different areas based on two independent checks: 1. **Size check**: Is the box physically large enough to be considered "Bulky"? 2. **Weight check**: Is the box heavy enough to be considered "Heavy"? These two properties are completely independent of each other. A tiny box of lead could be Heavy but not Bulky. A massive box of feathers could be Bulky but not Heavy. A giant crate of iron could be Both. An ordinary shoebox could be Neither. The key insight is that we don't need any complex logic — we simply evaluate both conditions separately, then combine the results to determine which of the four categories applies. approach: | We solve this using a **Direct Condition Evaluation** approach: **Step 1: Define the threshold constants** - `DIMENSION_THRESHOLD`: `10^4` (10,000) — any single dimension at or above this makes the box Bulky - `VOLUME_THRESHOLD`: `10^9` (1,000,000,000) — volume at or above this makes the box Bulky - `MASS_THRESHOLD`: `100` — mass at or above this makes the box Heavy   **Step 2: Check if the box is Bulky** - Calculate the volume: `length * width * height` - Check if any dimension >= `10^4` OR if volume >= `10^9` - Store result in a boolean `is_bulky`   **Step 3: Check if the box is Heavy** - Check if mass >= `100` - Store result in a boolean `is_heavy`   **Step 4: Return the appropriate category** - If both `is_bulky` and `is_heavy`: return `"Both"` - If `is_bulky` but not `is_heavy`: return `"Bulky"` - If `is_heavy` but not `is_bulky`: return `"Heavy"` - If neither: return `"Neither"` common_pitfalls: - title: Integer Overflow When Calculating Volume description: | The volume calculation `length * width * height` can produce very large numbers. With maximum dimensions of `10^5` each, the maximum volume is `10^15`. In languages with fixed-size integers (like 32-bit int in C++/Java), this will overflow since `2^31 - 1 ≈ 2.1 * 10^9`. Python handles arbitrary precision integers automatically, but in other languages you must use 64-bit integers (long) or check dimensions first before calculating volume. wrong_approach: "Using 32-bit integers for volume calculation" correct_approach: "Use 64-bit integers or check dimensions before volume" - title: Misreading the Thresholds description: | The problem uses scientific notation: `10^4` for dimensions and `10^9` for volume. It's easy to misread these or add/remove zeros. - Dimension threshold: 10,000 (ten thousand) - Volume threshold: 1,000,000,000 (one billion) Double-check your constants match these values exactly. wrong_approach: "Using 1000 instead of 10000 for dimension threshold" correct_approach: "Carefully verify thresholds: 10^4 = 10000, 10^9 = 1000000000" - title: Overcomplicating the Logic description: | With four possible outputs, it's tempting to write complex nested if-else chains. But since `is_bulky` and `is_heavy` are independent booleans, the logic is actually a simple 2x2 truth table. You can even simplify using string arrays or tuple lookups, but straightforward if-else works fine for this easy problem. wrong_approach: "Deeply nested conditionals checking all cases redundantly" correct_approach: "Evaluate both conditions independently, then combine" key_takeaways: - "**Decompose into independent checks**: When multiple conditions contribute to a result, evaluate each separately before combining" - "**Watch for overflow**: Volume and area calculations with large inputs can exceed 32-bit integer limits" - "**Truth table thinking**: With two boolean conditions and four outcomes, visualise as a 2x2 grid to ensure you cover all cases" - "**Constants over magic numbers**: Define thresholds as named constants for clarity and maintainability" time_complexity: "O(1). We perform a fixed number of comparisons and one multiplication, regardless of input values." space_complexity: "O(1). We only use a few boolean variables and store no data proportional to input." solutions: - approach_name: Direct Condition Evaluation is_optimal: true code: | def categorize_box(length: int, width: int, height: int, mass: int) -> str: # Define thresholds for classification DIMENSION_THRESHOLD = 10**4 # 10,000 VOLUME_THRESHOLD = 10**9 # 1,000,000,000 MASS_THRESHOLD = 100 # Calculate volume (Python handles large integers natively) volume = length * width * height # Check if box is Bulky: any dimension >= 10^4 OR volume >= 10^9 is_bulky = ( length >= DIMENSION_THRESHOLD or width >= DIMENSION_THRESHOLD or height >= DIMENSION_THRESHOLD or volume >= VOLUME_THRESHOLD ) # Check if box is Heavy: mass >= 100 is_heavy = mass >= MASS_THRESHOLD # Return category based on combination of conditions if is_bulky and is_heavy: return "Both" elif is_bulky: return "Bulky" elif is_heavy: return "Heavy" else: return "Neither" explanation: | **Time Complexity:** O(1) — Fixed number of operations regardless of input size. **Space Complexity:** O(1) — Only boolean variables and constants used. We evaluate the Bulky and Heavy conditions independently, then use simple conditionals to return the correct category. This approach is clean, readable, and efficient. - approach_name: Tuple Lookup is_optimal: false code: | def categorize_box(length: int, width: int, height: int, mass: int) -> str: # Calculate volume volume = length * width * height # Determine boolean flags is_bulky = ( length >= 10000 or width >= 10000 or height >= 10000 or volume >= 10**9 ) is_heavy = mass >= 100 # Use tuple as key to lookup result # (is_bulky, is_heavy) -> category categories = { (True, True): "Both", (True, False): "Bulky", (False, True): "Heavy", (False, False): "Neither" } return categories[(is_bulky, is_heavy)] explanation: | **Time Complexity:** O(1) — Dictionary lookup is constant time. **Space Complexity:** O(1) — Fixed-size dictionary with 4 entries. This approach uses a dictionary with tuple keys to map all four combinations directly to their results. While slightly more Pythonic and eliminates if-else chains, it's marginally less efficient due to dictionary lookup overhead. For this simple problem, the direct approach is preferred.