LeetCode 1277. Count Square Submatrices with All Ones

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题目

Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.

Example 1:

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Input: matrix =
[
  [0,1,1,1],
  [1,1,1,1],
  [0,1,1,1]
]
Output: 15
Explanation: 
There are 10 squares of side 1.
There are 4 squares of side 2.
There is  1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.

Example 2:

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Input: matrix = 
[
  [1,0,1],
  [1,1,0],
  [1,1,0]
]
Output: 7
Explanation: 
There are 6 squares of side 1.  
There is 1 square of side 2. 
Total number of squares = 6 + 1 = 7.

Constraints:

  • 1 <= arr.length <= 300
  • 1 <= arr[0].length <= 300
  • 0 <= arr[i][j] <= 1

思路

DP.

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dp[i][j] = edge of largest square with bottom right corner at (i, j)
dp[i][j] = min(dp[i – 1][j], dp[i – 1][j – 1], dp[i][j – 1]) + 1 if m[i][j] == 1 else 0
ans = sum(dp)
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Time complexity: O(n*m)
Space complexity: O(n*m)

代码

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class Solution(object):
    def countSquares(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: int
        """
        if matrix == []:
            return 0
        res = 0
        m, n = len(matrix), len(matrix[0])
        dp = [[0 for _ in range(n)] for _ in range(m)]
        for i in range(m):
            for j in range(n):
                dp[i][j] = matrix[i][j]
                if i and j and dp[i][j]:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
                res += dp[i][j]
        return res