# Leetcode Flood-Fill

#### 2022-07-15 09:01

> ##### Algorithms:
>
> #algorithm #BFS
>
> ##### Data structures:
>
> #DS
>
> ##### Difficulty:
>
> #coding_problem #difficulty_easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/flood-fill/)

---

### Problem

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 `color`. 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 `color`.

Return _the modified image after performing the flood fill_.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/06/01/flood1-grid.jpg)

```
**Input:** image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, color = 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, color = 0
**Output:** [[0,0,0],[0,0,0]]
**Explanation:** The starting pixel is already colored 0, so no changes are made to the image.
```

#### Constraints

- `m == image.length`
- `n == image[i].length`
- `1 <= m, n <= 50`
- `0 <= image[i][j], color < 216`
- `0 <= sr < m`
- `0 <= sc < n`

### Thoughts

> [!summary]
> This is a search problem, can be solved using DFS or BFS

This one can be optimized.

Initially, I wanted to use a hash map to record cells that are visited,
but the this takes up extra space.

Then I found out that I can use colors:

- if image[r][c] == color, this means
  - the cell is no need to correct(color == origcolor),
  - or this has been corrected
    so I don't have to go over again.

There are checks in the loop:

- check the color is not visited, as shown above
- check the coord is not OOB
- check the cell is equal to origColor, to only fill same origColor.

#TODO: Write in DFS

### Solution

```cpp
class Solution {
public:
  vector<vector<int>> floodFill(vector<vector<int>> &image, int sr, int sc,
                                int color) {
    int origColor = image[sr][sc];
    queue<pair<int, int>> todo;
    int m = image.size();
    int n = image[0].size();

    todo.push({sr, sc});
    int r, c;

    while (!todo.empty()) {
      r = todo.front().first;
      c = todo.front().second;
      todo.pop();

      if (image[r][c] != origColor || image[r][c] == color) {
        // already colored
        continue;
      }

      image[r][c] = color;

      if (r > 0) {
        todo.push({r - 1, c});
      }
      if (r < m - 1) {
        todo.push({r + 1, c});
      }
      if (c > 0) {
        todo.push({r, c - 1});
      }
      if (c < n - 1) {
        todo.push({r, c + 1});
      }
    }

    return image;
  }
};
```