How do you solve Count Artifacts That Can Be Extracted on LeetCode?

Count Artifacts That Can Be Extracted (LeetCode #2201) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set for quick lookups significantly reduces the time complexity.

What is the brute force solution for Count Artifacts That Can Be Extracted?

The brute force approach for Count Artifacts That Can Be Extracted is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute-force approach, we will check each artifact to see if all its parts have been excavated. This involves checking each excavation against all parts of each artifact, which can be slow but is straightforward. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Artifacts That Can Be Extracted?

Count Artifacts That Can Be Extracted uses the following concepts: Array, Hash Table, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Artifacts That Can Be Extracted?

Always consider the constraints and choose data structures that optimize for those. Explain your thought process clearly during the interview — it shows your understanding. Practice dry runs of your code with examples to ensure you understand the flow.

What are common mistakes when solving Count Artifacts That Can Be Extracted?

Not using a set for excavated cells, leading to inefficient lookups. Overcomplicating the logic by using nested loops unnecessarily.

#2201

Count Artifacts That Can Be Extracted

Medium

ArrayHash TableSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

In the optimal solution, we can utilize a set to quickly check if all parts of an artifact have been excavated. This reduces the need for nested loops and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a set of excavated cells from the dig array.
2Step 2: For each artifact, use a single loop to check if all its coordinates are in the excavated set.
3Step 3: Count the number of artifacts that can be extracted.

solution.py3 lines

1def countArtifacts(n, artifacts, dig):
2    excavated = set(tuple(d) for d in dig)
3    return sum(1 for r1, c1, r2, c2 in artifacts if all((r, c) in excavated for r in range(r1, r2 + 1) for c in range(c1, c2 + 1)))

ℹ

Complexity note: The time complexity is O(n) because we efficiently check each artifact against the excavated cells using a set. The space complexity remains O(n) for storing the excavated cells.

1Using a set for quick lookups significantly reduces the time complexity.
2Understanding the problem constraints helps in choosing the right data structures.

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