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countSquareSubMatrix.py
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countSquareSubMatrix.py
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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
@author: zenprog
Count Square Submatrices with All Ones
Given a m * n matrix of ones and zeros, return how many square submatrices have
all ones.
Example 1:
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:
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
"""
class Solution(object):
"""Object to test."""
def countSquares(self, matrix):
"""
Return how many square submatrices have all ones.
:type matrix: List[List[int]]
:rtype: int
"""
if not matrix:
return 0
result = 0
n, m = len(matrix), len(matrix[0])
for i in range(0, m):
result += matrix[n - 1][i]
for i in range(0, n):
result += matrix[i][m - 1]
result -= matrix[n - 1][m - 1]
for i in range(n - 2, -1, -1):
for j in range(m - 2, -1, -1):
if matrix[i][j] == 1:
matrix[i][j] = 1 + min(matrix[i + 1][j + 1],
matrix[i][j + 1], matrix[i + 1][j])
else:
matrix[i][j] = 0
result += matrix[i][j]
return result
obj = Solution()
matrix = [[0, 1, 1, 1], [1, 1, 1, 1], [0, 1, 1, 1]]
matrix1 = [[1, 0, 1], [1, 1, 0], [1, 1, 0]]
print(obj.countSquares(matrix))
print(obj.countSquares(matrix1))