Leetcode #661: Image Smoother

In this guide, we solve Leetcode #661 Image Smoother in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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).

Quick Facts

Difficulty: Easy
Premium: No
Tags: Array, Matrix

Intuition

Grid problems are easiest when you define clear row/column boundaries.

A consistent traversal order prevents off-by-one errors.

Approach

Iterate by rows, columns, or layers depending on the requirement.

Keep bounds updated as the traversal progresses.

Steps:

Define bounds or directions.
Visit cells in order.
Update result and move bounds.

Example

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

Python Solution

class Solution:
    def imageSmoother(self, img: List[List[int]]) -> List[List[int]]:
        m, n = len(img), len(img[0])
        ans = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                s = cnt = 0
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < m and 0 <= y < n:
                            cnt += 1
                            s += img[x][y]
                ans[i][j] = s // cnt
        return ans

Complexity

The time complexity is $O(m \times n)$ , where $m$ and $n$ are the number of rows and columns of $\textit{img}$ , respectively. The space complexity is $O(1)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

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).

Python Solution

class Solution:
    def imageSmoother(self, img: List[List[int]]) -> List[List[int]]:
        m, n = len(img), len(img[0])
        ans = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                s = cnt = 0
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < m and 0 <= y < n:
                            cnt += 1
                            s += img[x][y]
                ans[i][j] = s // cnt
        return ans

Leetcode #661: Image Smoother

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #661: Image Smoother

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview