LeetCode/1504. Count Submatrices With All Ones(Dynamic Programming) 20.7.6 Medium at master · BrianSong/LeetCode · GitHub

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Given a rows * columns matrix mat of ones and zeros, return how many submatrices have all ones.


Example 1:

Input: mat = [[1,0,1],
              [1,1,0],
              [1,1,0]]
Output: 13
Explanation:
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2.
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.
Example 2:

Input: mat = [[0,1,1,0],
              [0,1,1,1],
              [1,1,1,0]]
Output: 24
Explanation:
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3.
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2.
There are 2 rectangles of side 3x1.
There is 1 rectangle of side 3x2.
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
Example 3:

Input: mat = [[1,1,1,1,1,1]]
Output: 21
Example 4:

Input: mat = [[1,0,1],[0,1,0],[1,0,1]]
Output: 5


Constraints:

1 <= rows <= 150
1 <= columns <= 150
0 <= mat[i][j] <= 1


Solution: O(m * n * m) T and O(m * n) S ---------------- https://www.youtube.com/watch?time_continue=566&v=8HYXkNB39KA&feature=emb_logo
class Solution(object):
    def numSubmat(self, mat):
        """
        :type mat: List[List[int]]
        :rtype: int
        """
        if not mat or not mat[0]:
            return 0

        m, n = len(mat), len(mat[0])
        pre = [[0 for j in range(n)] for i in range(m)]
        for i in range(m):
            accumulate_one = 0
            for j in range(n - 1, -1, -1):
                if mat[i][j] == 1:
                    accumulate_one += 1
                else:
                    accumulate_one = 0
                pre[i][j] = accumulate_one
        res = 0
        for i in range(m):
            for j in range(n):
                min_value = float('inf')
                for k in range(i, m):
                    min_value = min(pre[k][j], min_value)
                    res += min_value

        return res