661 Image Smoother
Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
Example 1:
Input:
[[1,1,1],
[1,0,1],
[1,1,1]]
Output:
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
Explanation:
For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
For the point (1,1): floor(8/9) = floor(0.88888889) = 0
Note:
- 1.The value in the given matrix is in the range of [0, 255].
- 2.The length and width of the given matrix are in the range of [1, 150].
The Idea: Evolve a 3x3 filter along every element in the matrix. Sum the elements as you go, and count the number of active cells to avoid boundary issues.
Complexity: O(n*m) time and O(1) space
class Solution:
def imageSmoother(self, M):
"""
:type M: List[List[int]]
:rtype: List[List[int]]
"""
if not M:
return M
def is_valid(r, c):
return r >= 0 and c >= 0 and c < len(M[0]) and r < len(M)
n_rows = len(M)
n_cols = len(M[0])
M_new = [[0] * n_cols for _ in range(n_rows)]
rotations = [(1, 1), (-1, 1), (1, -1), (0, 1), (0, -1), (1, 0), (-1, 0), (-1, -1)]
for i in range(n_rows):
for j in range(0, n_cols):
n_active = 1
summ = M[i][j]
for x, y in rotations:
if is_valid(i + x, j + y):
n_active += 1
summ += M[i + x][j + y]
M_new[i][j] = summ // n_active
return M_new
Last modified 4yr ago