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Array
Matrix

661. Image Smoother

中文文档

Description

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

Solutions

Solution 1: Direct Traversal

We create a 2D array $\textit{ans}$ of size $m \times n$, where $\textit{ans}[i][j]$ represents the smoothed value of the cell in the $i$-th row and $j$-th column of the image.

For $\textit{ans}[i][j]$, we traverse the cell in the $i$-th row and $j$-th column of $\textit{img}$ and its surrounding 8 cells, calculate their sum $s$ and count $cnt$, then compute the average value $s / cnt$ and store it in $\textit{ans}[i][j]$.

After the traversal, we return $\textit{ans}$.

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

Python3

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

Java

class Solution {
    public int[][] imageSmoother(int[][] img) {
        int m = img.length;
        int n = img[0].length;
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int s = 0;
                int cnt = 0;
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x >= 0 && x < m && y >= 0 && y < n) {
                            ++cnt;
                            s += img[x][y];
                        }
                    }
                }
                ans[i][j] = s / cnt;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& img) {
        int m = img.size(), n = img[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 s = 0, cnt = 0;
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x < 0 || x >= m || y < 0 || y >= n) {
                            continue;
                        }
                        ++cnt;
                        s += img[x][y];
                    }
                }
                ans[i][j] = s / cnt;
            }
        }
        return ans;
    }
};

Go

func imageSmoother(img [][]int) [][]int {
	m, n := len(img), len(img[0])
	ans := make([][]int, m)
	for i, row := range img {
		ans[i] = make([]int, n)
		for j := range row {
			s, cnt := 0, 0
			for x := i - 1; x <= i+1; x++ {
				for y := j - 1; y <= j+1; y++ {
					if x >= 0 && x < m && y >= 0 && y < n {
						cnt++
						s += img[x][y]
					}
				}
			}
			ans[i][j] = s / cnt
		}
	}
	return ans
}

TypeScript

function imageSmoother(img: number[][]): number[][] {
    const m = img.length;
    const n = img[0].length;
    const ans: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            let s = 0;
            let cnt = 0;
            for (let x = i - 1; x <= i + 1; ++x) {
                for (let y = j - 1; y <= j + 1; ++y) {
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        ++cnt;
                        s += img[x][y];
                    }
                }
            }
            ans[i][j] = Math.floor(s / cnt);
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn image_smoother(img: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = img.len();
        let n = img[0].len();
        let mut ans = vec![vec![0; n]; m];
        for i in 0..m {
            for j in 0..n {
                let mut s = 0;
                let mut cnt = 0;
                for x in i.saturating_sub(1)..=(i + 1).min(m - 1) {
                    for y in j.saturating_sub(1)..=(j + 1).min(n - 1) {
                        s += img[x][y];
                        cnt += 1;
                    }
                }
                ans[i][j] = s / cnt;
            }
        }
        ans
    }
}