LeetCode #2201 — MEDIUM

Count Artifacts That Can Be Extracted

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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] = [r1_i, c1_i, r2_i, c2_i] denotes that the i^th artifact is buried in the subgrid where:

(r1_i, c1_i) is the coordinate of the top-left cell of the i^th artifact and
(r2_i, c2_i) is the coordinate of the bottom-right cell of the i^th artifact.

You will excavate some cells of the grid and remove all the mud from them. If the cell has a part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact are uncovered, you can extract it.

Given a 0-indexed 2D integer array dig where dig[i] = [r_i, c_i] indicates that you will excavate the cell (r_i, c_i), return the number of artifacts that you can extract.

The test cases are generated such that:

No two artifacts overlap.
Each artifact only covers at most 4 cells.
The entries of dig are unique.

Example 1:

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.

Example 2:

Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1],[1,1]]
Output: 2
Explanation: Both the red and blue artifacts have all parts uncovered (labeled with a 'D') and can be extracted, so we return 2.

Constraints:

1 <= n <= 1000
1 <= artifacts.length, dig.length <= min(n², 10⁵)
artifacts[i].length == 4
dig[i].length == 2
0 <= r1_i, c1_i, r2_i, c2_i, r_i, c_i <= n - 1
r1_i <= r2_i
c1_i <= c2_i
No two artifacts will overlap.
The number of cells covered by an artifact is at most 4.
The entries of dig are unique.

Step 01

Brute Force Baseline

Problem summary: 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. You will excavate some cells of the grid and remove all the mud from them. If the cell has a part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact are uncovered, you can extract it. Given a 0-indexed 2D integer array dig where dig[i] = [ri, ci] indicates that you will excavate the cell (ri, ci), return the number of artifacts that you can extract. The test cases are generated such that: No two artifacts overlap.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

2
[[0,0,0,0],[0,1,1,1]]
[[0,0],[0,1]]

Example 2

2
[[0,0,0,0],[0,1,1,1]]
[[0,0],[0,1],[1,1]]

Core Insight

What unlocks the optimal approach

Check if each coordinate of each artifact has been excavated. How can we do this quickly without iterating over the dig array every time?
Consider marking all excavated cells in a 2D boolean array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2201: Count Artifacts That Can Be Extracted
class Solution {
    private Set<Integer> s = new HashSet<>();
    private int n;

    public int digArtifacts(int n, int[][] artifacts, int[][] dig) {
        this.n = n;
        for (var p : dig) {
            s.add(p[0] * n + p[1]);
        }
        int ans = 0;
        for (var a : artifacts) {
            ans += check(a);
        }
        return ans;
    }

    private int check(int[] a) {
        int x1 = a[0], y1 = a[1], x2 = a[2], y2 = a[3];
        for (int x = x1; x <= x2; ++x) {
            for (int y = y1; y <= y2; ++y) {
                if (!s.contains(x * n + y)) {
                    return 0;
                }
            }
        }
        return 1;
    }
}

// Accepted solution for LeetCode #2201: Count Artifacts That Can Be Extracted
func digArtifacts(n int, artifacts [][]int, dig [][]int) (ans int) {
	s := map[int]bool{}
	for _, p := range dig {
		s[p[0]*n+p[1]] = true
	}
	check := func(a []int) int {
		x1, y1, x2, y2 := a[0], a[1], a[2], a[3]
		for x := x1; x <= x2; x++ {
			for y := y1; y <= y2; y++ {
				if !s[x*n+y] {
					return 0
				}
			}
		}
		return 1
	}
	for _, a := range artifacts {
		ans += check(a)
	}
	return
}

# Accepted solution for LeetCode #2201: Count Artifacts That Can Be Extracted
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)

// Accepted solution for LeetCode #2201: Count Artifacts That Can Be Extracted
use std::collections::HashSet;

impl Solution {
    pub fn dig_artifacts(n: i32, artifacts: Vec<Vec<i32>>, dig: Vec<Vec<i32>>) -> i32 {
        let mut s: HashSet<i32> = HashSet::new();
        for p in dig {
            s.insert(p[0] * n + p[1]);
        }
        let check = |a: &[i32]| -> i32 {
            let x1 = a[0];
            let y1 = a[1];
            let x2 = a[2];
            let y2 = a[3];
            for x in x1..=x2 {
                for y in y1..=y2 {
                    if !s.contains(&(x * n + y)) {
                        return 0;
                    }
                }
            }
            1
        };
        let mut ans = 0;
        for a in artifacts {
            ans += check(&a);
        }
        ans
    }
}

// Accepted solution for LeetCode #2201: Count Artifacts That Can Be Extracted
function digArtifacts(n: number, artifacts: number[][], dig: number[][]): number {
    const s: Set<number> = new Set();
    for (const [x, y] of dig) {
        s.add(x * n + y);
    }
    let ans = 0;
    const check = (a: number[]): number => {
        const [x1, y1, x2, y2] = a;
        for (let x = x1; x <= x2; ++x) {
            for (let y = y1; y <= y2; ++y) {
                if (!s.has(x * n + y)) {
                    return 0;
                }
            }
        }
        return 1;
    };
    for (const a of artifacts) {
        ans += check(a);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m + k)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.