> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/graphs-bfs-and-dfs/01-matrix.md).

# 01 Matrix

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

**Example 1:**

```
Input:
[[0,0,0],
 [0,1,0],
 [0,0,0]]

Output:
[[0,0,0],
 [0,1,0],
 [0,0,0]]
```

**Example 2:**

```
Input:
[[0,0,0],
 [0,1,0],
 [1,1,1]]

Output:
[[0,0,0],
 [0,1,0],
 [1,2,1]]
```

**Note:**

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

```java
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;
    }
}
```