221. Maximal Square

Leetcode: 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.

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-1:

class Solution {
    public int maximalSquare(char[][] matrix) {
        
        int rows = matrix.length;
        
        if(rows==0) return 0;
        
        int column = matrix[0].length;
        
        int [][] dp = new int[rows][column];
        int maxLength = 0;
        
        for(int i=0;i<rows;i++) {
            for(int j=0;j<column;j++) {
                if(matrix[i][j]=='1') {
                    dp[i][j] = 1;
                    maxLength = 1;
                }
            }
        }
        
        for(int i=1;i<rows;i++) {
            for(int j=1;j<column;j++) {
                if(dp[i][j]==0) continue;
                int min = Math.min(Math.min(dp[i-1][j-1], dp[i][j-1]), dp[i-1][j]) + 1;
                dp[i][j] = min;
                maxLength = Math.max(maxLength, min);
            }
        }
        
        return maxLength*maxLength;
    }
}

Solution-2:

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