Problem Description

You need to categorize a box based on its dimensions and mass. The box has four properties: length, width, height, and mass.

A box is classified as "Bulky" if it meets either of these conditions:

• Any single dimension (length, width, or height) is greater than or equal to 10,000
• The volume (length × width × height) is greater than or equal to 1,000,000,000 (10^9)

A box is classified as "Heavy" if:

• Its mass is greater than or equal to 100

Based on these classifications, the box falls into one of four categories:

• "Both" - if the box is both Bulky and Heavy
• "Bulky" - if the box is only Bulky (not Heavy)
• "Heavy" - if the box is only Heavy (not Bulky)
• "Neither" - if the box is neither Bulky nor Heavy

The solution uses bit manipulation to efficiently determine the category. It checks the bulky and heavy conditions, then combines them into an index using bitwise operations (heavy << 1 | bulky), which creates a 2-bit number representing all four possible states (00, 01, 10, 11). This index directly maps to the appropriate category string in the array ['Neither', 'Bulky', 'Heavy', 'Both'].

Intuition

The problem essentially asks us to classify a box into one of four categories based on two binary conditions: whether it's bulky and whether it's heavy. This immediately suggests we're dealing with a 2×2 decision matrix.

When we have two binary conditions, we get four possible combinations:

• Not bulky, not heavy → "Neither"
• Bulky, not heavy → "Bulky"
• Not bulky, heavy → "Heavy"
• Bulky, heavy → "Both"

The key insight is recognizing that these four states can be represented as a 2-bit number. If we treat heavy as the higher bit and bulky as the lower bit, we get:

• 00 (0 in decimal) → Neither condition is true
• 01 (1 in decimal) → Only bulky is true
• 10 (2 in decimal) → Only heavy is true
• 11 (3 in decimal) → Both conditions are true

This naturally maps to array indexing. By converting our boolean conditions to integers (0 or 1) and combining them using heavy << 1 | bulky, we create an index from 0 to 3. The left shift (<< 1) moves the heavy bit to the second position, and the OR operation (|) combines it with the bulky bit.

We can then create an array with the category strings in the exact order that corresponds to these indices: ['Neither', 'Bulky', 'Heavy', 'Both']. This eliminates the need for multiple if-else statements and makes the solution both elegant and efficient.

Learn more about Math patterns.

Solution Approach

The solution follows a simulation approach where we directly implement the problem's requirements.

Step 1: Calculate the volume

v = length * width * height

We compute the volume by multiplying all three dimensions.

Step 2: Determine if the box is bulky

bulky = int(any(x >= 10000 for x in (length, width, height)) or v >= 10**9)

This line checks two conditions for bulkiness:

• Uses any() to check if any dimension is greater than or equal to 10,000
• Checks if the volume is greater than or equal to 10^9
• The or operator ensures the box is bulky if either condition is true
• Converts the boolean result to an integer (0 for False, 1 for True)

Step 3: Determine if the box is heavy

heavy = int(mass >= 100)

Simply checks if the mass is at least 100 and converts the boolean to an integer.

Step 4: Create an index using bit manipulation

i = heavy << 1 | bulky

This combines the two binary values into a single index:

• heavy << 1 shifts the heavy bit left by 1 position (multiplies by 2)
• | bulky performs a bitwise OR with the bulky bit
• Results in: 0 (neither), 1 (bulky only), 2 (heavy only), or 3 (both)

Step 5: Return the corresponding category

d = ['Neither', 'Bulky', 'Heavy', 'Both']
return d[i]

The array is ordered to match the index values created in step 4, allowing direct lookup of the appropriate category string.

This approach elegantly handles all four cases without nested conditionals, making the code clean and efficient with O(1) time and space complexity.

Example Walkthrough

Let's walk through the solution with a concrete example to understand how the bit manipulation approach works.

Example: Consider a box with dimensions length = 2000, width = 3000, height = 1500, and mass = 150.

Step 1: Calculate the volume

v = 2000 × 3000 × 1500 = 9,000,000,000

Step 2: Determine if the box is bulky

• Check if any dimension ≥ 10,000:
  • 2000 < 10,000 ✗
  • 3000 < 10,000 ✗
  • 1500 < 10,000 ✗
• Check if volume ≥ 1,000,000,000:
  • 9,000,000,000 ≥ 1,000,000,000 ✓

The box is bulky! Converting to integer: bulky = 1

Step 3: Determine if the box is heavy

• Check if mass ≥ 100:
  • 150 ≥ 100 ✓

The box is heavy! Converting to integer: heavy = 1

Step 4: Create index using bit manipulation

i = heavy << 1 | bulky
i = 1 << 1 | 1
i = 2 | 1
i = 3

Breaking this down in binary:

• heavy = 1 in binary: 01
• Left shift by 1: 10 (which is 2 in decimal)
• bulky = 1 in binary: 01
• Bitwise OR: 10 | 01 = 11 (which is 3 in decimal)

Step 5: Return the corresponding category

d = ['Neither', 'Bulky', 'Heavy', 'Both']
d[3] = 'Both'

The function returns "Both" because the box is both bulky (due to its large volume) and heavy (mass ≥ 100).

This bit manipulation technique efficiently maps the four possible combinations of bulky/heavy states to array indices 0-3, eliminating the need for multiple conditional statements.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because:

• Checking if any dimension is >= 10000 involves at most 3 comparisons
• Computing the volume v = length * width * height is a constant-time operation with 2 multiplications
• Comparing volume with 10^9 is a single comparison
• Comparing mass with 100 is a single comparison
• The bitwise operations (<< and |) are constant-time
• Array indexing d[i] is constant-time
• All operations are performed a fixed number of times regardless of input values

The space complexity is O(1) because:

• The function uses a fixed number of variables: v, bulky, heavy, i, and d
• The list d always contains exactly 4 string elements, which is a constant amount of space
• No additional space is allocated that scales with the input
• All variables store single values (integers or references) that don't grow with input size

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Calculating Volume

The Problem: When calculating volume = length * width * height, the multiplication of three potentially large integers can cause integer overflow in languages with fixed-size integers. For example, if each dimension is close to 10,000, the volume would be around 10^12, which exceeds the range of 32-bit integers.

Why It Happens:

• In languages like C++ or Java, multiplying large integers without proper type casting can lead to overflow
• Even in Python (which handles arbitrary precision), extremely large numbers could cause performance issues

Solution:

# Instead of checking volume >= 10^9 after calculation,
# check if the product would exceed the threshold during calculation
def is_volume_bulky(length, width, height):
    # Check if length * width would already exceed threshold
    if length >= 10**9 // (width * height) if width * height != 0 else False:
        return True
    return length * width * height >= 10**9

# Alternative: Use early termination
def is_volume_bulky(length, width, height):
    threshold = 10**9
    # If any dimension is >= threshold^(1/3), volume will likely exceed threshold
    cube_root = 1000  # approximately 10^9^(1/3)
    if length >= cube_root and width >= cube_root and height >= cube_root:
        return True
    return length * width * height >= 10**9

2. Incorrect Operator Precedence in Bit Manipulation

The Problem: When using the bit manipulation approach heavy << 1 | bulky, developers might misunderstand the operator precedence or incorrectly implement the logic.

Why It Happens:

• Confusion about whether << or | has higher precedence
• Swapping the positions of heavy and bulky bits

Solution:

# Be explicit with parentheses for clarity
index = (heavy << 1) | bulky

# Or use multiplication which is more intuitive
index = heavy * 2 + bulky

# Ensure the array matches the bit pattern:
# 00 (binary) = 0 = Neither (not heavy, not bulky)
# 01 (binary) = 1 = Bulky (not heavy, bulky)
# 10 (binary) = 2 = Heavy (heavy, not bulky)
# 11 (binary) = 3 = Both (heavy, bulky)
categories = ['Neither', 'Bulky', 'Heavy', 'Both']

3. Off-by-One Errors in Threshold Comparisons

The Problem: Using > instead of >= or vice versa when checking thresholds.

Why It Happens:

• Misreading the problem requirements
• Confusion about whether the threshold values are inclusive or exclusive

Solution:

# Correct: use >= for all threshold checks as per problem statement
is_bulky = any(d >= 10000 for d in (length, width, height)) or volume >= 10**9
is_heavy = mass >= 100

# Incorrect examples to avoid:
# is_bulky = any(d > 10000 for d in dimensions)  # Wrong!
# is_heavy = mass > 100  # Wrong!