General idea is `BFS`

. Some small tricks:

- At beginning, set cell value to
`Integer.MAX_VALUE`

if it is not`0`

. - If newly calculated distance
`>=`

current distance, then we don't need to explore that cell again.

```
public class Solution {
public List<List<Integer>> updateMatrix(List<List<Integer>> matrix) {
int m = matrix.size();
int n = matrix.get(0).size();
Queue<int[]> queue = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix.get(i).get(j) == 0) {
queue.offer(new int[] {i, j});
}
else {
matrix.get(i).set(j, Integer.MAX_VALUE);
}
}
}
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!queue.isEmpty()) {
int[] cell = queue.poll();
for (int[] d : dirs) {
int r = cell[0] + d[0];
int c = cell[1] + d[1];
if (r < 0 || r >= m || c < 0 || c >= n ||
matrix.get(r).get(c) <= matrix.get(cell[0]).get(cell[1]) + 1) continue;
queue.add(new int[] {r, c});
matrix.get(r).set(c, matrix.get(cell[0]).get(cell[1]) + 1);
}
}
return matrix;
}
}
```

LeetCode has changed the function signature. Updated code:

```
public class Solution {
public int[][] updateMatrix(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
Queue<int[]> queue = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 0) {
queue.offer(new int[] {i, j});
}
else {
matrix[i][j] = Integer.MAX_VALUE;
}
}
}
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!queue.isEmpty()) {
int[] cell = queue.poll();
for (int[] d : dirs) {
int r = cell[0] + d[0];
int c = cell[1] + d[1];
if (r < 0 || r >= m || c < 0 || c >= n ||
matrix[r][c] <= matrix[cell[0]][cell[1]] + 1) continue;
queue.add(new int[] {r, c});
matrix[r][c] = matrix[cell[0]][cell[1]] + 1;
}
}
return matrix;
}
}
```