Introduction

In this blog post, we will explore how to solve the LeetCode problem "2201" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. 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. 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 <= r1i, c1i, r2i, c2i, ri, ci <= n - 1 r1i <= r2i c1i <= c2i No two artifacts will overlap. The number of cells covered by an artifact is at most 4. The entries of dig are unique.

Explanation

Here's a breakdown of the approach, complexity, and the Python code for the problem:

• High-Level Approach:

  • Represent the dug cells using a set for efficient lookup.
  • Iterate through each artifact and check if all its cells are present in the dug cells set.
  • If all cells of an artifact are dug, increment the extracted artifact count.

• Complexity:

  • Runtime Complexity: O(len(artifacts) * 4 + len(dig)), which simplifies to O(len(artifacts) + len(dig)) because each artifact covers at most 4 cells.
  • Storage Complexity: O(len(dig)) for storing the dug cells in a set.

• Python Code:

Code

    def dig_artifats(n: int, artifats: list[list[int]], dig: list[list[int]]) -> int:    """
    Calculates the number of artifacts that can be extracted after digging.

    Args:
        n: The size of the grid.
        artifacts: A list of artifact coordinates [r1, c1, r2, c2].
        dig: A list of dug cell coordinates [r, c].

    Returns:
        The number of extracted artifacts.
    """

    dug_cells = set((r, c) for r, c in dig)
    extracted_artifacts = 0

    for r1, c1, r2, c2 in artifacts:
        extracted = True
        for r in range(r1, r2 + 1):
            for c in range(c1, c2 + 1):
                if (r, c) not in dug_cells:
                    extracted = False
                    break
            if not extracted:
                break
        if extracted:
            extracted_artifacts += 1

    return extracted_artifacts