• Time:O(mn)
• Space:O(mn)

## 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;
}
};
``````

## 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;
}
}
``````

## 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
``````