打砖块

/* 
* Solution approach using Java Streams: 
* Similar to the previous solution but leveraging Java Streams for iteration and collection operations. 
*/

import java.util.*;
import java.util.stream.*;

public class SolutionStream {
    public int[] hitBricks(int[][] grid, int[][] hits) {
        int m = grid.length, n = grid[0].length;
        Arrays.stream(hits).forEach(hit -> grid[hit[0]][hit[1]] -= 1);
        IntStream.range(0, n).forEach(j -> dfs(grid, 0, j));
        int[] res = new int[hits.length];
        IntStream.range(0, hits.length).map(i -> hits.length - 1 - i).forEach(i -> {
            int r = hits[i][0], c = hits[i][1];
            grid[r][c] += 1;
            if (grid[r][c] == 1 && isConnectedToTop(grid, r, c)) {
                res[i] = dfs(grid, r, c) - 1;
            }
        });
        return res;
    }

    private int dfs(int[][] grid, int r, int c) {
        if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length || grid[r][c] != 1) {
            return 0;
        }
        grid[r][c] = 2;
        return 1 + IntStream.of(new int[]{1, -1}).map(d -> dfs(grid, r + d, c) + dfs(grid, r, c + d)).sum();
    }

    private boolean isConnectedToTop(int[][] grid, int r, int c) {
        if (r == 0) {
            return true;
        }
        return Stream.of(new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}).anyMatch(dir -> {
            int x = r + dir[0], y = c + dir[1];
            return x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && grid[x][y] == 2;
        });
    }
}