Leetcode: Maximal Square

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 0

Return 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;
    }
}

Update on 4/2/19: Rolling array

public class Solution {
    /**
     * @param matrix: a matrix of 0 and 1
     * @return: an integer
     */
    public int maxSquare(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return 0;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        
        int maxLen = 0;
        
        int[][] dp = new int[2][n];
        for (int i = 0; i < n; i++) {
            if (matrix[0][i] == 1) {
                dp[0][i] = 1;
                maxLen = Math.max(maxLen, dp[0][i]);
            }
        }
        
        int cur = 0;
        int old = 0;
        
        for (int i = 1; i < m; i++) {
            old = cur;
            cur = 1 - cur;
            dp[cur][0] = matrix[i][0] == 1 ? 1 : 0;
            maxLen = Math.max(maxLen, dp[cur][0]);
            
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == 1) {
                    dp[cur][j] = Math.min(dp[old][j - 1], Math.min(dp[old][j], dp[cur][j - 1])) + 1;
                } else {
                    dp[cur][j] = 0;
                }
                
                maxLen = Math.max(maxLen, dp[cur][j]);
            }
        }
        
        return maxLen * maxLen;
        
    }
}