221. Maximal Square

/*
 * @lc app=leetcode id=221 lang=cpp
 *
 * [221] Maximal Square
 *
 * https://leetcode.com/problems/maximal-square/description/
 *
 * algorithms
 * Medium (47.66%)
 * Likes:    10401
 * Dislikes: 237
 * Total Accepted:    759.3K
 * Total Submissions: 1.6M
 * Testcase Example:  '[["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]'
 *
 * Given an m x n binary matrix filled with 0's and 1's, find the largest
 * square containing only 1's and return its area.
 *
 *
 * Example 1:
 *
 *
 * Input: matrix =
 * [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
 * Output: 4
 *
 *
 * Example 2:
 *
 *
 * Input: matrix = [["0","1"],["1","0"]]
 * Output: 1
 *
 *
 * Example 3:
 *
 *
 * Input: matrix = [["0"]]
 * Output: 0
 *
 *
 *
 * Constraints:
 *
 *
 * m == matrix.length
 * n == matrix[i].length
 * 1 <= m, n <= 300
 * matrix[i][j] is '0' or '1'.
 *
 *
 */

// @lc code=start
class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix)
    {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dp(m, vector<int>(n));
        int d = 0;
        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] == '1')
                {
                    dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
                }
                d = max(d, dp[i][j]);
            }
        }
        return d * d;
    }
};
// @lc code=end

• T: $O(m \cdot n)$
• S: $O(m \cdot n)$

2 - Space Improved