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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.
1public class Solution {
2 public int maximalSquare(char[][] matrix) {
3 if (matrix.length == 0) return 0;
4 int m = matrix.length, n = matrix[0].length;
5 int[][] dp = new int[m+1][n+1];
6 int maxSide = 0;
7 for (int i = 1; i <= m; i++) {
8 for (int j = 1; j <= n; j++) {
9 if (matrix[i-1][j-1] == '1') {
10 dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
11 maxSide = Math.max(maxSide, dp[i][j]);
12 }
13 }
14 }
15 return maxSide * maxSide;
16 }
17}The Java solution utilizes a 2D integer array for the DP table, initialized with extra size for boundary conditions. The nested loops go over the matrix, using Math.min and Math.max to update the DP cells and track the largest square area.
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 public int MaximalSquare(char[][] matrix) {
if (matrix.Length == 0) return 0;
int m = matrix.Length, n = matrix[0].Length;
int[] prev = new int[n + 1];
int[] current = new int[n + 1];
int maxSide = 0;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (matrix[i - 1][j - 1] == '1') {
current[j] = Math.Min(Math.Min(prev[j], current[j - 1]), prev[j - 1]) + 1;
maxSide = Math.Max(maxSide, current[j]);
}
}
Array.Copy(current, prev, n + 1);
Array.Clear(current, 0, n + 1);
}
return maxSide * maxSide;
}
}C# utilizes a similar logic with single dimensional `int` arrays for processing, swapping them with each iteration over rows. Utilizing Array.Copy and Array.Clear to manage elements over iterations.