Image Smoother Python Solution with Tests

Tested Python solution for LeetCode 661 with 14 pytest cases. Generate a practice environment with lcpy.

LeetCode 661, Easy. Topics: Array, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 14 parametrized pytest cases, and a playground notebook:

lcpy gen -n 661   # by problem number
lcpy gen -s image_smoother   # by problem name

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Problem

An image smoother is a filter of the size 3 x 3 that can be applied to each cell of an image by rounding down the average of the cell and the eight surrounding cells (i.e., the average of the nine cells in the blue smoother). If one or more of the surrounding cells of a cell is not present, we do not consider it in the average (i.e., the average of the four cells in the red smoother).

Given an m x n integer matrix img representing the grayscale of an image, return the image after applying the smoother on each cell of it.

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Examples

Input: img = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[0,0,0],[0,0,0],[0,0,0]]
Explanation: For the points (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0.
For the points (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.

Input: img = [[100,200,100],[200,50,200],[100,200,100]]
Output: [[137,141,137],[141,138,141],[137,141,137]]
Explanation: For the points (0,0), (0,2), (2,0), (2,2): floor((100+200+200+50)/4) = floor(137.5) = 137.
For the points (0,1), (1,0), (1,2), (2,1): floor((200+200+50+200+100+100)/6) = floor(141.666667) = 141.
For the point (1,1): floor((50+200+200+200+200+100+100+100+100)/9) = floor(138.888889) = 138.

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Constraints

• m == img.length
• n == img[i].length
• 1 <= m, n <= 200
• 0 <= img[i][j] <= 255

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(m * n)
    def image_smoother(self, img: list[list[int]]) -> list[list[int]]:
        m, n = len(img), len(img[0])
        result: list[list[int]] = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                total = count = 0
                for x in range(max(0, i - 1), min(m, i + 2)):
                    for y in range(max(0, j - 1), min(n, j + 2)):
                        total += img[x][y]
                        count += 1
                result[i][j] = total // count
        return result

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Complexity

Time	Space
O(m * n)	O(m * n)

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Tags

NeetCode All.