[LeetCode] Problem 661 - Image Smoother

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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

Input:

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[[1,1,1],
 [1,0,1],
 [1,1,1]]

Output:

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[[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].

Code

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private int[][] dirs = {{-1, -1}, {0, -1}, {1, -1}, {-1, 0}, {0, 0}, {1, 0}, {-1, 1}, {0, 1}, {1, 1}};

public int[][] imageSmoother(int[][] M) {
    int m = M.length;
    int n = M[0].length;

    int[][] result = new int[m][n];

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++)
            result[i][j] = smooth(M, m, n, i, j);
    }

    return result;
}

private int smooth(int[][] M, int m, int n, int i, int j) {
    int count = 0;
    int sum = 0;

    for (int[] dir : dirs) {
        int x = i + dir[0];
        int y = j + dir[1];

        if (x < 0 || y < 0 || x >= m || y >= n)
            continue;
        
        count++;
        sum += M[x][y];
    }

    return sum / count;
}