Sponsored
Sponsored
In this approach, a dynamic programming (DP) table is used to store the size of the largest square whose bottom-right corner is at each cell. For each cell (i, j) with a value of '1', we check its top (i-1, j), left (i, j-1), and top-left (i-1, j-1) neighbors to determine the size of the largest square ending at (i, j). If any of these neighbors are '0', the square cannot extend to include (i, j). The formula is: dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1. The maximum value in the DP table will be the side length of the largest square, and the area is its square.
Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
Space Complexity: O(m * n), due to the DP table.
1class Solution:
2 def maximalSquare(self, matrix):
3 if not matrix:
4 return 0
5 m, n = len(matrix), len(matrix[0])
6 dp = [[0] * (n + 1) for _ in range(m + 1)]
7 max_side = 0
8 for i in range(1, m + 1):
9 for j in range(1, n + 1):
10 if matrix[i-1][j-1] == '1':
11 dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
12 max_side = max(max_side, dp[i][j])
13 return max_side * max_sideThe Python solution initializes a DP table with one extra row and column filled with zeros, to handle boundary conditions without additional checks. It iterates through the matrix, updating the DP table using the given formula and tracking the maximum square side found.
This approach uses a similar DP strategy but optimizes space by utilizing a one-dimensional array instead of a full 2D DP table. The key idea is that while processing the matrix row by row, previous rows' information will be partially redundant. Hence, we can maintain only the current and previous row data in separate arrays or even use a single array with swap states.
Time Complexity: O(m * n)
Space Complexity: O(n)
1#
The C code uses two dynamically allocated arrays as buffers for current and previous row calculations. After each row processing, a swap is conducted, and the current array is reset to zeros. It uses only linear space, which improves the space efficiency.