# 2022-05-21

## [733. Flood Fill](https://leetcode.com/problems/flood-fill/)

### Description

An image is represented by an `m x n` integer grid `image` where `image[i][j]` represents the pixel value of the image.

You are also given three integers `sr`, `sc`, and `newColor`. You should perform a **flood fill** on the image starting from the pixel `image[sr][sc]`.

To perform a **flood fill**, consider the starting pixel, plus any pixels connected **4-directionally** to the starting pixel of the same color as the starting pixel, plus any pixels connected **4-directionally** to those pixels (also with the same color), and so on. Replace the color of all of the aforementioned pixels with `newColor`.

Return *the modified image after performing the flood fill*.

**Example 1:**

![](https://img.content.cc/a/2022/05/21/12-18-09-325-4ab3dd289ed8562296ab9c4ec0218c1d-c93280.jpeg)

```
Input: image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, newColor = 2
Output: [[2,2,2],[2,2,0],[2,0,1]]
Explanation: From the center of the image with position (sr, sc) = (1, 1) (i.e., the red pixel), all pixels connected by a path of the same color as the starting pixel (i.e., the blue pixels) are colored with the new color.
Note the bottom corner is not colored 2, because it is not 4-directionally connected to the starting pixel.
```

**Example 2:**

```
Input: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, newColor = 2
Output: [[2,2,2],[2,2,2]]
```

**Constraints:**

* `m == image.length`
* `n == image[i].length`
* `1 <= m, n <= 50`
* `0 <= image[i][j], newColor < 2^16`
* `0 <= sr < m`
* `0 <= sc < n`

### Solution

#### Approach #0: DFS

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

func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
    if image[sr][sc] != newColor {
        originColor := image[sr][sc]
        dfs(image, sr, sc, originColor, newColor)
    }
    return image
}

func dfs(image [][]int, sr int, sc int, originColor, newColor int) {
    if image[sr][sc] == originColor {
        image[sr][sc] = newColor
        for i := 0; i < 4; i++ {
            newSr, newSc := sr+dx[i], sc+dy[i]
            if newSr >= 0 && newSr < len(image) && newSc >= 0 && newSc < len(image[0]) {
                dfs(image, newSr, newSc, originColor, newColor)
            }
        }
    }
}
```

#### Approach #1: BFS

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

func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
    if image[sr][sc] == newColor {
        return image
    }
    originColor := image[sr][sc]
    image[sr][sc] = newColor
    queue := [][]int{[]int{sr, sc}}
    for i := 0; i < len(queue); i++ {
        for j := 0; j < 4; j++ {
            x, y := queue[i][0]+dx[j], queue[i][1]+dy[j]
            if x >= 0 && x < len(image) && y >= 0 && y < len(image[0]) && image[x][y] == originColor {
                image[x][y] = newColor
                queue = append(queue, []int{x, y})

            }
        }
    }
    return image
}

```

## [695. Max Area of Island](https://leetcode.com/problems/max-area-of-island/)

### Description

You are given an `m x n` binary matrix `grid`. An island is a group of `1`'s (representing land) connected **4-directionally** (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The **area** of an island is the number of cells with a value `1` in the island.

Return *the maximum **area** of an island in* `grid`. If there is no island, return `0`.

**Example 1:**

![](https://img.content.cc/a/2022/05/21/14-04-44-545-c462d022e77d90bc42de4f86e1ed4e9b-75ba3e.jpeg)

```
Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
```

**Example 2:**

```
Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0
```

**Constraints:**

* `m == grid.length`
* `n == grid[i].length`
* `1 <= m, n <= 50`
* `grid[i][j]` is either `0` or `1`.

### Solution

#### Approach #0: BFS

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

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

    res := 0
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == 0 {
                continue
            }
            grid[i][j] = 0
            queue := [][]int{{i, j}}
            area := 1
            for ii := 0; ii < len(queue); ii++ {
                cell := queue[ii]
                for jj := 0; jj < 4; jj++ {
                    x, y := cell[0]+dx[jj], cell[1]+dy[jj]
                    if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0 {
                        grid[x][y] = 0
                        area++
                        queue = append(queue, []int{x, y})
                    }
                }
            }
            res = max(res, area)
        }
    }
    return res
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}
```

#### Approach #1: DFS (**Recursive**)

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

func maxAreaOfIsland(grid [][]int) int {
    if len(grid) == 0 {
        return 0
    }

    res := 0
    for i := 0; i < len(grid); i++ {
        for j := 0; j < len(grid[0]); j++ {
            if grid[i][j] == 0 {
                continue
            }
            area := dfs(grid, i, j)
            res = max(res, area)
        }
    }
    return res
}

func dfs(grid [][]int, x, y int) int {
    grid[x][y] = 0
    res := 1
    for i := 0; i < 4; i++ {
        xx, yy := x+dx[i], y+dy[i]
        if xx >= 0 && xx < len(grid) && yy >= 0 && yy < len(grid[0]) && grid[xx][yy] > 0 {
            res = res + dfs(grid, xx, yy)
        }
    }
    return res
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}
```

#### Approach #2: DFS + Stack (**Iterative**)

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

func maxAreaOfIsland(grid [][]int) int {
    if len(grid) == 0 {
        return 0
    }

    res := 0
    for i := 0; i < len(grid); i++ {
        for j := 0; j < len(grid[0]); j++ {
            s := [][]int{{i, j}}
            area := 0
            for len(s) > 0 {
                cell := s[len(s)-1]
                s = s[:len(s)-1]
                x, y := cell[0], cell[1]
                if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) || grid[x][y] == 0 {
                    continue
                }
                grid[x][y] = 0
                area++
                for i := 0; i < 4; i++ {
                    xx, yy := x+dx[i], y+dy[i]
                    s = append(s, []int{xx, yy})
                }

            }
            res = max(res, area)
        }
    }
    return res
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}
```