Leetcode #2201: Count Artifacts That Can Be Extracted
In this guide, we solve Leetcode #2201 Count Artifacts That Can Be Extracted in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an n x n 0-indexed grid with some artifacts buried in it. You are given the integer n and a 0-indexed 2D integer array artifacts describing the positions of the rectangular artifacts where artifacts[i] = [r1i, c1i, r2i, c2i] denotes that the ith artifact is buried in the subgrid where: (r1i, c1i) is the coordinate of the top-left cell of the ith artifact and (r2i, c2i) is the coordinate of the bottom-right cell of the ith artifact.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Simulation
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1]]
Output: 1
Explanation:
The different colors represent different artifacts. Excavated cells are labeled with a 'D' in the grid.
There is 1 artifact that can be extracted, namely the red artifact.
The blue artifact has one part in cell (1,1) which remains uncovered, so we cannot extract it.
Thus, we return 1.
Python Solution
class Solution:
def digArtifacts(
self, n: int, artifacts: List[List[int]], dig: List[List[int]]
) -> int:
def check(a: List[int]) -> bool:
x1, y1, x2, y2 = a
return all(
(x, y) in s for x in range(x1, x2 + 1) for y in range(y1, y2 + 1)
)
s = {(i, j) for i, j in dig}
return sum(check(a) for a in artifacts)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.