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

Example:

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

 

 

Solution:

Dynamic Programming.

When matrix[i][j] is ‘1’,

dp[i][j] = min(matrix[i -1][j],

matrix[i][j – 1],

matrix[i – 1][j -1]) + 1.

The maximal square will be the maximum dp[i][j] square.

 

class Solution {
    public int maximalSquare(char[][] matrix) {
        int n = matrix.length;
        if (n == 0) {
            return 0;
        }
        int m = matrix[0].length;
        if (m == 0) {
            return 0;
        }
        
        int[][] cmat = new int[n][m];
        final int INF = Integer.MAX_VALUE;
        int maxv = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (matrix[i][j] == '0') {
                    cmat[i][j] = 0;
                } else {
                    int minv = INF;
                    if (i > 0) {
                        minv = Math.min(minv, cmat[i-1][j]);
                    } else {
                        minv = 0;
                    }
                    if (j > 0) {
                        minv = Math.min(minv, cmat[i][j-1]);
                    } else {
                        minv = 0;
                    }
                    if (i > 0 && j > 0) {
                        minv = Math.min(minv, cmat[i-1][j-1]);
                    } else {
                        minv = 0;
                    }
                    cmat[i][j] = minv + 1;
                    maxv = Math.max(maxv, cmat[i][j]);
                }
            }
        }
        return maxv * maxv;
    }
}

 

Time complexity: O(n).

Space complexity: O(n).