221. Maximal Square

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



Solution:

Method 1: Dynamic Programming

dp[i][j]: max size to form a square from this point as lower right point.

dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1], dp[i][j - 1]) + 1



Code:

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

Method 2: Optimized with rolling array



Code:

public class Solution {
    public int maximalSquare(char[][] matrix) {
        int max = 0;
        if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) {
            return 0;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] dp = new int[2][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || j == 0) {
                    dp[i % 2][j] = matrix[i][j] - '0';
                }
                else if (matrix[i][j] == '0') {
                    dp[i % 2][j] = 0;
                }
                else {
                    dp[i % 2][j] = Math.min(Math.min(dp[(i - 1) % 2][j], dp[i % 2][j - 1]), dp[(i - 1) % 2][j - 1]) + 1;
                }
                max = Math.max(max, dp[i % 2][j]);
            }
        }
        return max * max;
    }
}