Formatted question description: https://leetcode.ca/all/661.html

661. Image Smoother (Easy)

Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.

Example 1:

Input:
[[1,1,1],
 [1,0,1],
 [1,1,1]]
Output:
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]
Explanation:
For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
For the point (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

Note:

  1. The value in the given matrix is in the range of [0, 255].
  2. The length and width of the given matrix are in the range of [1, 150].

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/image-smoother/
// Time: O(MN)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> imageSmoother(vector<vector<int>>& A) {
        int M = A.size(), N = A[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 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;
                        ans[i][j] += A[x][y];
                        ++cnt;
                    }
                }
                ans[i][j] /= cnt;
            }
        }
        return ans;
    }
};

Java

  • class Solution {
        public int[][] imageSmoother(int[][] M) {
            if (M == null || M.length == 0 || M[0].length == 0)
                return M;
            int rows = M.length, columns = M[0].length;
            int[][] smooth = new int[rows][columns];
            int[][] directions = { {-1, -1}, {-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1} };
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < columns; j++) {
                    int sum = M[i][j];
                    int count = 1;
                    for (int[] direction : directions) {
                        int newRow = i + direction[0], newColumn = j + direction[1];
                        if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns) {
                            sum += M[newRow][newColumn];
                            count++;
                        }
                    }
                    smooth[i][j] = sum / count;
                }
            }
            return smooth;
        }
    }
    
  • // OJ: https://leetcode.com/problems/image-smoother/
    // Time: O(MN)
    // Space: O(1)
    class Solution {
    public:
        vector<vector<int>> imageSmoother(vector<vector<int>>& A) {
            int M = A.size(), N = A[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 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;
                            ans[i][j] += A[x][y];
                            ++cnt;
                        }
                    }
                    ans[i][j] /= cnt;
                }
            }
            return ans;
        }
    };
    
  • class Solution(object):
      def imageSmoother(self, M):
        """
        :type M: List[List[int]]
        :rtype: List[List[int]]
        """
        m = len(M)
        n = len(M[0])
        ans = [[0] * n for _ in range(m)]
    
        for i in range(m):
          for j in range(n):
            cnt = 0
            sums = 0
            for di in range(-1, 2):
              for dj in range(-1, 2):
                newi, newj = i + di, j + dj
                if 0 <= newi < m and 0 <= newj < n:
                  cnt += 1
                  sums += M[newi][newj]
            ans[i][j] = sums / cnt
        return ans
    
    

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