# 2022-05-23

## [542. 01 Matrix](https://leetcode.com/problems/01-matrix/)

### Description

Given an `m x n` binary matrix `mat`, return *the distance of the nearest* `0` *for each cell*.

The distance between two adjacent cells is `1`.

**Example 1:**

![](https://img.content.cc/a/2022/05/23/19-29-42-885-5362838c3aa004246f109b39f73f2703-6e6211.jpeg)

```
Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]
```

**Example 2:**

![](https://img.content.cc/a/2022/05/23/19-30-11-789-e52361d105f13159883dd2340e186d13-a94720.jpeg)

```
Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]
```

**Constraints:**

* `m == mat.length`
* `n == mat[i].length`
* `1 <= m, n <= 10^4`
* `1 <= m * n <= 10^4`
* `mat[i][j]` is either `0` or `1`.
* There is at least one `0` in `mat`.

### Solution

#### Approach #0: BFS (Time Limit Exceeded) ❌

```go
var (
    dx = []int{0, 1, 0, -1}
    dy = []int{1, 0, -1, 0}
)

func updateMatrix(mat [][]int) [][]int {
    if len(mat) == 0 {
        return mat
    }
    m, n := len(mat), len(mat[0])

    res := make([][]int, m)
    for i := 0; i < m; i++ {
        res[i] = make([]int, n)
    }

    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            queue := [][]int{{i, j}}
            depth := -1
            t := make([][]int, m)
            for i := 0; i < m; i++ {
                t[i] = make([]int, n)
            }
            for len(queue) > 0 {
                size := len(queue)
                for k := 0; k < size; k++ {
                    x, y := queue[k][0], queue[k][1]
                    t[x][y] = 1
                    if mat[x][y] == 0 {
                        queue = queue[:0]
                        break
                    } else {
                        for p := 0; p < 4; p++ {
                            xx, yy := x+dx[p], y+dy[p]
                            if xx >= 0 && xx < m && yy >= 0 && yy < n && t[xx][yy] != 1 {
                                queue = append(queue, []int{xx, yy})
                            }
                        }
                    }
                }
                depth++
                if len(queue) > 0 {
                    queue = queue[size:]
                }

            }
            res[i][j] = depth
        }
    }
    return res
}
```

#### Approach #1: BFS

```go
var (
    dx = []int{0, 1, 0, -1}
    dy = []int{1, 0, -1, 0}
)

func updateMatrix(mat [][]int) [][]int {
    if len(mat) == 0 {
        return mat
    }
    m, n := len(mat), len(mat[0])

    res := make([][]int, m)
    checked := make([][]int, m)
    for i := 0; i < m; i++ {
        res[i] = make([]int, n)
        checked[i] = make([]int, n)
    }

    var queue [][]int
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if mat[i][j] == 0 {
                queue = append(queue, []int{i, j})
                checked[i][j] = 1
            }
        }
    }

    depth := 0
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            x, y := queue[i][0], queue[i][1]
            res[x][y] = depth
            for p := 0; p < 4; p++ {
                xx, yy := x+dx[p], y+dy[p]
                if xx >= 0 && xx < m && yy >= 0 && yy < n && checked[xx][yy] != 1 {
                    checked[xx][yy] = 1
                    queue = append(queue, []int{xx, yy})
                }
            }
        }
        depth++
        queue = queue[size:]
    }

    return res
}
```

## [994. Rotting Oranges](https://leetcode.com/problems/rotting-oranges/)

### Description

You are given an `m x n` `grid` where each cell can have one of three values:

* `0` representing an empty cell,
* `1` representing a fresh orange, or
* `2` representing a rotten orange.

Every minute, any fresh orange that is **4-directionally adjacent** to a rotten orange becomes rotten.

Return *the minimum number of minutes that must elapse until no cell has a fresh orange*. If *this is impossible, return* `-1`.

**Example 1:**

![](https://img.content.cc/a/2022/05/23/19-30-36-297-461a69dd506a0a1b5d67b4ecfd535eb2-1ce1e5.png)

```
Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
```

**Example 2:**

```
Input: grid = [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
```

**Example 3:**

```
Input: grid = [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
```

**Constraints:**

* `m == grid.length`
* `n == grid[i].length`
* `1 <= m, n <= 10`
* `grid[i][j]` is `0`, `1`, or `2`.

### Solution

#### Approach #0: BFS

```go
var (
    dx = []int{0, 1, 0, -1}
    dy = []int{1, 0, -1, 0}
)

func orangesRotting(grid [][]int) int {
    if len(grid) == 0 {
        return 0
    }
    m, n := len(grid), len(grid[0])
    all1 := 0
    var queue [][]int
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == 1 {
                all1++
            }
            if grid[i][j] == 2 {
                queue = append(queue, []int{i, j})
            }
        }
    }
    if len(queue) == 0 && all1 == 0 {
        return 0
    }

    depth := -1
    count1 := -len(queue)
    for len(queue) > 0 {
        size := len(queue)
        for i := 0; i < size; i++ {
            x, y := queue[i][0], queue[i][1]
            count1++
            for j := 0; j < 4; j++ {
                xx, yy := x+dx[j], y+dy[j]
                if xx >= 0 && xx < m && yy >= 0 && yy < n && grid[xx][yy] == 1 {
                    grid[xx][yy] = 2
                    queue = append(queue, []int{xx, yy})
                }
            }

        }
        fmt.Println(all1, count1)
        depth++
        queue = queue[size:]
    }
    if all1 > count1 {
        return -1
    }
    return depth
}
```


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