661. Image Smoother LeetCode Solution

Last updated: February 19, 2025

In this guide, you will get 661. Image Smoother LeetCode Solution with the best time and space complexity. The solution to Image Smoother problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Image Smoother solution in C++
Image Smoother solution in Java
Image Smoother solution in Python
Additional Resources

661. Image Smoother LeetCode Solution image

Problem Statement of Image Smoother

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

Complexity Analysis

Time Complexity: O(mn)
Space Complexity: O(mn)

661. Image Smoother LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
    const int m = M.size();
    const int n = M[0].size();
    vector<vector<int>> ans(m, vector<int>(n));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int ones = 0;
        int count = 0;
        for (int x = max(0, i - 1); x < min(m, i + 2); ++x)
          for (int y = max(0, j - 1); y < min(n, j + 2); ++y) {
            ones += M[x][y];
            ++count;
          }
        ans[i][j] = ones / count;
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

661. Image Smoother LeetCode Solution in Java

class Solution {
  public int[][] imageSmoother(int[][] M) {
    final int m = M.length;
    final int n = M[0].length;
    int ans[][] = new int[m][n];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int ones = 0;
        int count = 0;
        for (int y = Math.max(0, i - 1); y < Math.min(m, i + 2); ++y)
          for (int x = Math.max(0, j - 1); x < Math.min(n, j + 2); ++x) {
            ones += M[y][x];
            ++count;
          }
        ans[i][j] = ones / count;
      }

    return ans;
  }
}
// code provided by PROGIEZ

661. Image Smoother LeetCode Solution in Python

class Solution:
  def imageSmoother(self, M: list[list[int]]) -> list[list[int]]:
    m = len(M)
    n = len(M[0])
    ans = [[0 for j in range(n)] for i in range(m)]

    for i in range(m):
      for j in range(n):
        ones = 0
        count = 0
        for y in range(max(0, i - 1), min(m, i + 2)):
          for x in range(max(0, j - 1), min(n, j + 2)):
            ones += M[y][x]
            count += 1
        ans[i][j] = ones // count

    return ans
 # code by PROGIEZ

Additional Resources

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