How do you solve Stamping the Grid on LeetCode?

Stamping the Grid (LeetCode #2132) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Using prefix sums can significantly reduce the number of checks needed.

What is the brute force solution for Stamping the Grid?

The brute force approach for Stamping the Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries to place the stamp at every possible position in the grid and checks if it can cover all empty cells. This method is straightforward but inefficient, especially for larger grids. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Stamping the Grid?

Stamping the Grid uses the following concepts: Array, Greedy, Matrix, Prefix Sum. The recommended patterns to study are: Greedy, Prefix Sum, Matrix.

What are the interview tips for Stamping the Grid?

Always clarify the problem constraints and edge cases before diving into coding. Discuss your thought process and any potential optimizations with the interviewer. Practice explaining your code as you write it to simulate the interview environment.

What are common mistakes when solving Stamping the Grid?

Failing to account for edge cases where the stamp size exceeds grid dimensions. Not properly checking for overlaps with occupied cells.

#2132

Stamping the Grid

Hard

ArrayGreedyMatrixPrefix SumGreedyPrefix SumMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses a greedy approach combined with prefix sums to efficiently determine if all empty cells can be covered by stamps. This method reduces the number of checks needed by leveraging cumulative sums.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to track the number of empty cells in any submatrix of size stampHeight x stampWidth.
2Step 2: Iterate through the grid and for each empty cell, check if placing a stamp at that position would cover all empty cells in the stamp area.
3Step 3: If all empty cells can be covered, return true; otherwise, return false.

solution.py13 lines

1def canStamp(grid, stampHeight, stampWidth):
2    m, n = len(grid), len(grid[0])
3    prefix = [[0] * (n + 1) for _ in range(m + 1)]
4    
5    for i in range(m):
6        for j in range(n):
7            prefix[i + 1][j + 1] = grid[i][j] + prefix[i][j + 1] + prefix[i + 1][j] - prefix[i][j]
8    
9    for i in range(m - stampHeight + 1):
10        for j in range(n - stampWidth + 1):
11            if prefix[i + stampHeight][j + stampWidth] - prefix[i][j + stampWidth] - prefix[i + stampHeight][j] + prefix[i][j] == 0:
12                return True
13    return False

ℹ

Complexity note: The time complexity is O(m * n) because we need to compute the prefix sums for the entire grid. The space complexity is also O(m * n) for storing the prefix sum array.

1Using prefix sums can significantly reduce the number of checks needed.
2Greedy approaches can often lead to optimal solutions when combined with cumulative data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.