Introduction

In this blog post, we will explore how to solve the LeetCode problem "661" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the image smoother problem:

• Iterate through each cell of the input image img.
• For each cell, calculate the average of the cell and its surrounding 8 neighbors, handling boundary conditions (cells on the edge/corners).

• Store the smoothed value (floor of the average) in the corresponding cell of the output image.

• Runtime Complexity: O(m*n), where m is the number of rows and n is the number of columns in the image. Storage Complexity: O(m*n) due to the creation of the result matrix.

Code

    def imageSmoother(img: list[list[int]]) -> list[list[int]]:
    m = len(img)
    n = len(img[0])
    result = [[0] * n for _ in range(m)]

    for i in range(m):
        for j in range(n):
            total = 0
            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