Find the maximum size of square of side ‘X’ in a matrix

We use dynamic programming to solve this.
Space complexity = O(2*n)
Time complexity = O(n^3).

public class maxSquareX {
    public static void main(String[] args) {

        char[][] input = { 
                    { 'X', '0', 'X', 'X', 'X' },
                    { 'X', 'X', 'X', 'X', 'X' },
                    { 'X', 'X', '0', 'X', '0' },
                    { 'X', 'X', 'X', 'X', 'X' },
                    { 'X', 'X', 'X', '0', '0' },
                  };

        int[][] rows = new int[input.length][input[0].length];
        int[][] cols = new int[input.length][input[0].length];

        for (int i = 0; i < input.length; i++) {
            for (int j = 0; j < input[0].length; j++) {
                if (input[i][j] == '0') {
                    rows[i][j] = cols[i][j] = 0;
                } else {
                    rows[i][j] = (i == 0) ? 1 : rows[i - 1][j] + 1;
                    cols[i][j] = (j == 0) ? 1 : cols[i][j - 1] + 1;
                }
            }
        }
        int max = 1; // since this is the minimum answer we can get int small; 
        // Start from bottom right 
        for (int i = input.length - 1; i >= 0; i--) {
            for (int j = input[0].length - 1; j >= 0; j--) {
                small = Math.min(rows[i][j], cols[i][j]);
                while (small > max) {
                    // we use index [j - small + 1] because small might be
                    // input.length + 1 and when that is subtracted from 'j' we
                    // might get an 'ArrayIndexOutOfBoundsException'.
                    if (small <= rows[i][j - small + 1] && small <= cols[i - small + 1][j]) {
                        max = small;
                    } else
                        small--;
                }
            }
        }
    }
}

Output

3