LeetCode Solutions

661. Image Smoother

Time: $O(mn)$

Space: $O(mn)$

			

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;
  }
};
			

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;
  }
}
			

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