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.

Example 1:

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

Example 2:

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

Constraints:

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

Problem Analysis:

• Let m be the number of rows and n be the number of columns in the input image.
• The algorithm iterates through each pixel in the image, visiting each pixel constant times.
• The nested loops for finding the sum and count of pixel intensities have a constant number of iterations (a 3x3 neighborhood).
• Thus, the time complexity is O(m×n).
• The space complexity is O(m×n) since the algorithm creates a new 2D array to store the smoothed image.

Solutions:

class Solution:
    def imageSmoother(self, img: List[List[int]]) -> List[List[int]]:
        rows, cols = len(img), len(img[0])
        res = [[0] * cols for _ in range(rows)]

        for i in range(rows):
            for j in range(cols):
                total_sum, count = 0, 0

                for k in range(max(0, i-1), min(rows, i+2)):
                    for l in range(max(0, j-1), min(cols, j+2)):
                        total_sum += img[k][l]
                        count += 1

                res[i][j] = total_sum // count

        return res

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