2201. Count Artifacts That Can Be Extracted

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.

• simulation solution

class Solution {
    void marking(int id, vector<vector<int>>& matrix, vector<int>& arti, vector<int>& artiSize) {
        for(int i = arti[0]; i <= arti[2]; i++)
            for(int j = arti[1]; j <= arti[3]; j++) {
                artiSize[id]++;
                matrix[i][j] = id;
            }
    }
public:
    int digArtifacts(int n, vector<vector<int>>& artifacts, vector<vector<int>>& dig) {
        vector<vector<int>> matrix(n, vector<int>(n,0));
        vector<int> artiSize(artifacts.size() + 1, 0);
        for(int i = 0; i < artifacts.size(); i++) {
            marking(i + 1, matrix, artifacts[i], artiSize);
        }
        for(auto d : dig) {
            artiSize[matrix[d[0]][d[1]]]--;
            matrix[d[0]][d[1]] = 0;
        }
        int res = 0;
        for(int i = 1; i < artiSize.size(); i++) {
            if(artiSize[i] == 0) res++;
        }
        return res;
    }
};

• hash set solution

class Solution {
public:
    int digArtifacts(int n, vector<vector<int>>& artifacts, vector<vector<int>>& dig) {
        unordered_set<int> us;
        for(auto d : dig) {
            us.insert(d[0] * 1000 + d[1]);
        }
        int res = 0;
        for(auto arti : artifacts) {
            bool find = true;
            for (int i = arti[0]; i <= arti[2] and find; i++) {
                for (int j = arti[1]; j <= arti[3] and find; j++) {
                    if (!us.count(i * 1000 + j)) find = false;
                }
            }
            if(find) res++;
        }
        return res;
    }
};