Problem Description

You have a ship with an n x n cargo deck, where each cell can hold exactly one container. Each container weighs exactly w units.

The ship has a maximum weight capacity of maxWeight units that it can safely carry.

You need to determine the maximum number of containers you can load onto the ship without exceeding the weight limit.

The solution approach recognizes that:

• The deck has n × n total cells available
• Each container weighs w units
• The total weight cannot exceed maxWeight

Therefore, the maximum number of containers is limited by either:

1. The physical space on the deck: n × n containers
2. The weight constraint: maxWeight ÷ w containers (rounded down)

The answer is the smaller of these two values, which can be calculated as min(n × n × w, maxWeight) ÷ w.

Intuition

The key insight is recognizing that we have two constraints limiting how many containers we can load:

1. Physical space constraint: The deck has n × n cells, so we can't load more than n × n containers regardless of weight.

2. Weight constraint: The ship can carry at most maxWeight units of weight. Since each container weighs w units, we can load at most maxWeight ÷ w containers (rounded down).

We need to satisfy both constraints simultaneously, so we take the minimum of these two limits.

To find this minimum elegantly, we can think of it as:

• The maximum possible weight if we fill the entire deck is n × n × w
• But we can't exceed maxWeight, so the actual weight we can load is min(n × n × w, maxWeight)
• To get the number of containers from this weight, we divide by w

This gives us the formula: min(n × n × w, maxWeight) ÷ w

This approach works because dividing the minimum allowable weight by the weight per container gives us exactly the maximum number of containers we can load while respecting both the space and weight limitations.

Learn more about Math patterns.

Solution Approach

The implementation is straightforward and uses pure mathematics without any complex data structures or algorithms.

The solution calculates the maximum number of containers in a single expression:

return min(n * n * w, maxWeight) // w

Breaking down the implementation:

1. Calculate total possible weight: n * n * w computes the weight if we were to fill every cell on the n × n deck with a container of weight w.

2. Apply weight constraint: min(n * n * w, maxWeight) ensures we don't exceed the ship's maximum weight capacity. This gives us the actual weight we can load.

3. Convert weight to container count: We use integer division // w to convert the total weight back to the number of containers. Since each container weighs exactly w units, dividing the total weight by w gives us the container count.

The beauty of this solution is its efficiency - it runs in O(1) time complexity and uses O(1) space complexity, as it performs just a few arithmetic operations regardless of the input size.

No loops, recursion, or additional data structures are needed. The mathematical formula directly computes the answer by considering both the physical and weight constraints simultaneously.

Example Walkthrough

Let's walk through an example with n = 3, w = 10, and maxWeight = 50.

Step 1: Identify the constraints

• The deck has 3 × 3 = 9 cells total
• Each container weighs 10 units
• The ship can carry at most 50 units of weight

Step 2: Calculate the maximum based on physical space

• If we fill all cells: 9 containers
• Total weight would be: 9 × 10 = 90 units

Step 3: Calculate the maximum based on weight limit

• With maxWeight = 50 and w = 10
• Maximum containers by weight: 50 ÷ 10 = 5 containers

Step 4: Apply the formula Using min(n × n × w, maxWeight) ÷ w:

• n × n × w = 3 × 3 × 10 = 90
• min(90, 50) = 50
• 50 ÷ 10 = 5

Result: 5 containers

The weight constraint is the limiting factor here. Even though the deck has space for 9 containers, we can only load 5 containers (weighing 50 units total) without exceeding the ship's weight capacity.

Let's verify with another example where space is the limiting factor:

• n = 2, w = 5, maxWeight = 100
• Space allows: 2 × 2 = 4 containers
• Weight allows: 100 ÷ 5 = 20 containers
• Using the formula: min(2 × 2 × 5, 100) ÷ 5 = min(20, 100) ÷ 5 = 20 ÷ 5 = 4
• Result: 4 containers (limited by deck space)

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a fixed number of arithmetic operations regardless of the input size. It calculates n * n * w, compares it with maxWeight using the min() function, and then performs integer division by w. All these operations take constant time.

The space complexity is O(1) because the function only uses a constant amount of extra space. No additional data structures are created, and the space used does not depend on the input values n, w, or maxWeight.

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

Common Pitfalls

1. Integer Overflow in Large Input Cases

The most significant pitfall in this solution is the potential for integer overflow when calculating n * n * w. If n and w are large values, their product could exceed the maximum integer limit in some programming languages.

Example scenario:

• If n = 10^5 and w = 10^5, then n * n * w = 10^15, which could cause overflow in languages with 32-bit integers.

Solution: Instead of calculating the full product first, rearrange the logic to avoid large intermediate values:

def maxContainers(self, n: int, w: int, maxWeight: int) -> int:
    # Check if weight constraint is the limiting factor first
    max_by_weight = maxWeight // w
    max_by_space = n * n
  
    # Return the minimum of the two constraints
    return min(max_by_weight, max_by_space)

2. Division by Zero

If w = 0, the code will throw a division by zero error. While this might be outside the problem constraints, defensive programming suggests handling this edge case.

Solution: Add a guard clause at the beginning:

def maxContainers(self, n: int, w: int, maxWeight: int) -> int:
    if w == 0:
        return 0  # No containers can be loaded if they weigh nothing
  
    return min(n * n * w, maxWeight) // w

3. Confusion About Problem Interpretation

The code comment mentions "n containers available" but the actual calculation uses n * n, suggesting an n × n grid. This inconsistency could lead to misunderstanding and incorrect modifications.

Solution: Ensure comments accurately reflect the problem:

def maxContainers(self, n: int, w: int, maxWeight: int) -> int:
    """
    Calculate the maximum number of containers that can be loaded on an n×n deck.
  
    Args:
        n: Dimension of the square deck (n×n grid)
        w: Weight of each container
        maxWeight: Maximum total weight capacity
    """
    # The deck has n×n cells, each can hold one container
    return min(n * n * w, maxWeight) // w