How do you solve Maximum Containers on a Ship on LeetCode?

Maximum Containers on a Ship (LeetCode #3492) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Total cells are n squared.

What is the time complexity of Maximum Containers on a Ship?

The optimal solution for Maximum Containers on a Ship runs in O(1) time and uses O(1) space. Direct calculations lead to O(1) complexity since we don't iterate through cells.

What is the brute force solution for Maximum Containers on a Ship?

The brute force approach for Maximum Containers on a Ship is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each cell on the deck to see if we can place a container without exceeding the weight limit. This involves iterating through all possible cells. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Maximum Containers on a Ship?

Maximum Containers on a Ship uses the following concepts: Math. The recommended patterns to study are: Greedy, Mathematical Optimization.

What are the interview tips for Maximum Containers on a Ship?

Explain your thought process clearly. Discuss edge cases like maxWeight being less than w. Practice similar problems to build confidence.

What are common mistakes when solving Maximum Containers on a Ship?

Forgetting to check both total cells and weight limits. Using integer division incorrectly.

#3492

Maximum Containers on a Ship

Easy

MathGreedyMathematical Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

Instead of checking each cell, we can directly calculate the maximum number of containers based on the weight limit and the grid size.

⚙️

Algorithm

3 steps

1Step 1: Calculate total cells as n * n.
2Step 2: Calculate maximum containers as maxWeight / w.
3Step 3: Return the minimum of total cells and maximum containers.

solution.py4 lines

1def maxContainers(n, w, maxWeight):
2    total_cells = n * n
3    max_containers = maxWeight // w
4    return min(total_cells, max_containers)

ℹ

Complexity note: Direct calculations lead to O(1) complexity since we don't iterate through cells.

1Total cells are n squared.
2Weight limits dictate the maximum number of containers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.