Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0Return 4.

**Understand the problem:**

This question can be solved by using DP.

-- Define dp[i][j] as the length of the maximal square of which the right bottom point ended with matrix[i][j].

-- Initial value dp[0][j] = matrix[0][j]; dp[i][0] = matrix[i][0];

-- Transit function: If matrix[i][j] == 1, dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1;

-- Final state, max(dp[i][j] * dp[i][j])

**Code (Java):**

public class Solution { public int maximalSquare(char[][] matrix) { if (matrix == null || matrix.length == 0) { return 0; } int rows = matrix.length; int cols = matrix[0].length; int[][] dp = new int[rows][cols]; // Initialization for (int i = 0; i < cols; i++) { dp[0][i] = matrix[0][i] - '0'; } for (int i = 0; i < rows; i++) { dp[i][0] = matrix[i][0] - '0'; } for (int i = 1; i < rows; i++) { for (int j = 1; j < cols; j++) { if (matrix[i][j] == '1') { dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1; } } } int maxArea = 0; for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { maxArea = Math.max(maxArea, dp[i][j] * dp[i][j]); } } return maxArea; } }

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