Leetcode #2132: Stamping the Grid
In this guide, we solve Leetcode #2132 Stamping the Grid in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Array, Matrix, Prefix Sum
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
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, 1):
for j, v in enumerate(row, 1):
s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + v
d = [[0] * (n + 2) for _ in range(m + 2)]
for i in range(1, m - stampHeight + 2):
for j in range(1, n - stampWidth + 2):
x, y = i + stampHeight - 1, j + stampWidth - 1
if s[x][y] - s[x][j - 1] - s[i - 1][y] + s[i - 1][j - 1] == 0:
d[i][j] += 1
d[i][y + 1] -= 1
d[x + 1][j] -= 1
d[x + 1][y + 1] += 1
for i, row in enumerate(grid, 1):
for j, v in enumerate(row, 1):
d[i][j] += d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1]
if v == 0 and d[i][j] == 0:
return False
return True
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.