221. Maximal Square

https://leetcode.com/problems/maximal-square/

solution

• 2d动态规划, dp[i][j]以matrix[i-1][j-1]为右下底的全是1的最大square的边长

• 因为涉及到前面的状态，i-1和j-1来简化初始化过程

• Space can be optimized as we don't need to keep the whole dp grid as we progress down the rows in matrix

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        if not matrix:
            return 0

        m = len(matrix)
        n = len(matrix[0])

        dp = [[0] * (n + 1) for _ in range(m + 1)]

        max_len = 0
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == '1':
                    # 有些tricky
                    dp[i+1][j+1] = min(dp[i][j], dp[i+1][j], dp[i][j+1]) + 1
                    if dp[i+1][j+1] > max_len:
                        max_len = dp[i+1][j+1]
        return max_len * max_len

时间复杂度：O(mn) 空间复杂度：OnmN)

follow up

85. Maximal Rectangle