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Formatted question description: https://leetcode.ca/all/2132.html
2132. Stamping the Grid
- Difficulty: Hard.
- Related Topics: Array, Greedy, Matrix, Prefix Sum.
- Similar Questions: Maximal Square, Bomb Enemy, Matrix Block Sum.
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 <= 105
-
1 <= m * n <= 2 * 105
-
grid[r][c]
is either0
or1
. -
1 <= stampHeight, stampWidth <= 105
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; } }
-
Todo
-
class Solution: def possibleToStamp( self, grid: List[List[int]], stampHeight: int, stampWidth: int ) -> bool: m, n = len(grid), len(grid[0]) s = [[0] * (n + 1) for _ in range(m + 1)] for i, row in enumerate(grid): for j, v in enumerate(row): s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + v d = [[0] * (n + 1) for _ in range(m + 1)] for i, row in enumerate(grid): for j, v in enumerate(row): if v == 0: x, y = i + stampHeight, j + stampWidth if x <= m and y <= n and s[x][y] - s[x][j] - s[i][y] + s[i][j] == 0: d[i][j] += 1 d[i][y] -= 1 d[x][j] -= 1 d[x][y] += 1 cnt = [[0] * (n + 1) for _ in range(m + 1)] for i, row in enumerate(grid): for j, v in enumerate(row): cnt[i + 1][j + 1] = cnt[i + 1][j] + cnt[i][j + 1] - cnt[i][j] + d[i][j] if v == 0 and cnt[i + 1][j + 1] == 0: return False return True ############ # 2132. Stamping the Grid # https://leetcode.com/problems/stamping-the-grid/ class Solution: def possibleToStamp(self, grid: List[List[int]], H: int, W: int) -> bool: rows, cols = len(grid), len(grid[0]) prefix = [[0] * (cols + 1) for _ in range(rows + 1)] for x in range(rows): for y in range(cols): prefix[x + 1][y + 1] = prefix[x + 1][y] + prefix[x][y + 1] - prefix[x][y] + grid[x][y] diff = [[0] * (cols + 1) for _ in range(rows + 1)] for x in range(rows - H + 1): for y in range(cols - W + 1): rect = prefix[x + H][y + W] - prefix[x + H][y] - prefix[x][y + W] + prefix[x][y] if rect == 0: diff[x][y] += 1 diff[x + H][y] -= 1 diff[x][y + W] -= 1 diff[x + H][y + W] += 1 for x in range(rows + 1): for y in range(cols): diff[x][y + 1] += diff[x][y] for y in range(cols + 1): for x in range(rows): diff[x + 1][y] += diff[x][y] for x in range(rows): for y in range(cols): if grid[x][y] == 0 and diff[x][y] == 0: return False return True
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).