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.

Example

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

Return4.

public int maxSquare(int[][] matrix) {
    if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return 0;
    }
    int m = matrix.length, n = matrix[0].length;
    int[][] dp = new int[m][n];
    int edge = 0;
    for(int i = 0; i < m; i++) {
        dp[i][0] = matrix[i][0];
        edge = Math.max(edge, dp[i][0]);
    }
    for(int j = 0; j < n; j++) {
        dp[0][j] = matrix[0][j];
        edge = Math.max(edge, dp[0][j]);
    }
    for(int i = 1; i < m; i++) {
        for(int j = 1; j < n; j++) {
            if(matrix[i][j] != 0) {
                dp[i][j] = Math.min(dp[i][j - 1], Math.min(dp[i - 1][j], dp[i - 1][j - 1])) + 1;
                edge = Math.max(edge, dp[i][j]);
            }
        }
    }
    return (int)(edge * edge);
}

Rolling array optimization:

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