2132. Stamping the Grid

Problem

You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied).

You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements:

• Cover all the empty cells.

• Do not cover any of the occupied cells.

• We can put as many stamps as we want.

• Stamps can overlap with each other.

• Stamps are not allowed to be rotated.

• Stamps must stay completely inside the grid.

Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false.

  Example 1:

Input: grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3
Output: true
Explanation: We have two overlapping stamps (labeled 1 and 2 in the image) that are able to cover all the empty cells.

Example 2:

Input: grid = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]], stampHeight = 2, stampWidth = 2 
Output: false 
Explanation: There is no way to fit the stamps onto all the empty cells without the stamps going outside the grid.

  Constraints:

• m == grid.length

• n == grid[r].length

• 1 <= m, n <= 10^5

• 1 <= m * n <= 2 * 10^5

• grid[r][c] is either 0 or 1.

• 1 <= stampHeight, stampWidth <= 10^5

Solution

class Solution {
    private boolean canPaved(int[][] grid, int is, int js, int ie, int je) {
        for (int i = is; i <= ie; i++) {
            for (int j = js; j <= je; j++) {
                if (grid[i][j] == 1) {
                    return true;
                }
            }
        }
        return false;
    }

    public boolean possibleToStamp(int[][] grid, int h, int w) {
        int rl = grid[0].length;
        for (int i = 0; i < grid.length; i++) {
            int[] row = grid[i];
            int prev = -1;
            for (int j = 0; j < row.length; j++) {
                if (row[j] == 0) {
                    if (j + 1 < rl
                            && row[j + 1] == 1
                            && j - w + 1 >= 0
                            && i + 1 < grid.length
                            && grid[i + 1][j] == 1
                            && i - h + 1 >= 0
                            && canPaved(grid, i - h + 1, j - w + 1, i, j)) {
                        return false;
                    }
                    continue;
                }
                if (1 < j - prev && j - prev <= w) {
                    return false;
                }
                prev = j;
            }
            if (1 < row.length - prev && row.length - prev <= w) {
                return false;
            }
        }
        for (int i = 0; i < rl; i++) {
            int prev = -1;
            for (int j = 0; j < grid.length; j++) {
                if (grid[j][i] == 0) {
                    continue;
                }
                if (1 < j - prev && j - prev <= h) {
                    return false;
                }
                prev = j;
            }
            if (1 < grid.length - prev && grid.length - prev <= h) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).