# Flood Fill
**Difficulty:** EASY
[External](https://leetcode.com/problems/flood-fill)
Canonical: https://scaleengineer.com/dsa/problems/flood-fill
**Algorithms:** [Depth-First Search](https://scaleengineer.com/algorithms/depth-first-search), [Breadth-First Search](https://scaleengineer.com/algorithms/breadth-first-search)
**Data structures:** Array, Matrix
**Companies:** [Criteo](https://scaleengineer.com/companies/criteo), [Palantir Technologies](https://scaleengineer.com/companies/palantir-technologies), [Akuna Capital](https://scaleengineer.com/companies/akuna-capital), [Sumo Logic](https://scaleengineer.com/companies/sumo-logic), [OpenAI](https://scaleengineer.com/companies/openai)
---
## Problem
You are given an image represented by an `m x n` grid of integers `image`, where `image[i][j]` represents the pixel value of the image. You are also given three integers `sr`, `sc`, and `color`. Your task is to perform a **flood fill** on the image starting from the pixel `image[sr][sc]`.

To perform a **flood fill**:

1. Begin with the starting pixel and change its color to `color`.
2. Perform the same process for each pixel that is **directly adjacent** (pixels that share a side with the original pixel, either horizontally or vertically) and shares the **same color** as the starting pixel.
3. Keep **repeating** this process by checking neighboring pixels of the _updated_ pixels and modifying their color if it matches the original color of the starting pixel.
4. The process **stops** when there are **no more** adjacent pixels of the original color to update.

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

**Example 1:**

**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:**

![](https://assets.glich.co/dsa/flood-fill/image0.jpg)

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 horizontally or vertically 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 with 0, which is the same as the target color. Therefore, 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`

# Approaches
## Depth-First Search (Recursive)
This approach utilizes recursion to perform a Depth-First Search (DFS) starting from the given pixel. It explores as far as possible along each branch before backtracking. The system's call stack is implicitly used to keep track of the pixels to visit.
**Time:** `O(N * M)`, where `N` is the number of rows and `M` is the number of columns. In the worst-case scenario, we might have to visit every pixel in the grid. · **Space:** `O(N * M)` in the worst case. This is due to the recursion depth of the call stack. If the area to be filled forms a long, winding path, the stack depth could be proportional to the number of pixels in the grid.
**Pros:** The recursive implementation is often very intuitive and leads to concise code for graph traversal problems.; It's easy to reason about the logic.
**Cons:** For large grids or deeply connected components, this approach can lead to a `StackOverflowError` if the recursion depth exceeds the system's stack limit.
### Explanation
The core idea is to have a recursive function that explores the grid. First, we check an important edge case: if the starting pixel's color is already the target color, no action is needed, and we can return the image immediately. This prevents infinite recursion. We store the original color of the starting pixel. We then call a recursive helper function, let's call it `dfs`. The `dfs` function takes the image, current coordinates (row, col), the original color, and the new color as arguments.

**Base Cases for recursion:** The function returns if the current coordinates are out of the grid's bounds, or if the color of the pixel at the current coordinates is not the original color we are trying to replace.

**Recursive Step:** If the base cases are not met, we update the color of the current pixel to the new color. Then, we make four recursive calls for the adjacent pixels (up, down, left, and right) to continue the flood fill process.

```java
class Solution {
    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int initialColor = image[sr][sc];
        if (initialColor != color) {
            dfs(image, sr, sc, initialColor, color);
        }
        return image;
    }

    private void dfs(int[][] image, int r, int c, int initialColor, int newColor) {
        if (r < 0 || r >= image.length || c < 0 || c >= image[0].length || image[r][c] != initialColor) {
            return;
        }
        image[r][c] = newColor;
        dfs(image, r + 1, c, initialColor, newColor);
        dfs(image, r - 1, c, initialColor, newColor);
        dfs(image, r, c + 1, initialColor, newColor);
        dfs(image, r, c - 1, initialColor, newColor);
    }
}
```
### Algorithm
- Get the color of the starting pixel `(sr, sc)`, let's call it `initialColor`.
- If `initialColor` is the same as the target `color`, return the image as no changes are needed. This check is crucial to prevent infinite recursion.
- Create a recursive helper function `dfs(image, r, c, initialColor, color)`.
- Inside `dfs`:
    - Check for boundary conditions: if `r` or `c` is out of bounds, return.
    - Check for color condition: if `image[r][c]` is not equal to `initialColor`, return.
    - If the conditions are passed, update the pixel's color: `image[r][c] = color`.
    - Make recursive calls for the four neighbors: `dfs(r+1, c)`, `dfs(r-1, c)`, `dfs(r, c+1)`, `dfs(r, c-1)`.
- Start the process by calling `dfs(image, sr, sc, initialColor, color)`.
- Return the modified `image`.

## Breadth-First Search (Iterative with Queue)
This approach uses an iterative Breadth-First Search (BFS) algorithm with a queue data structure. It explores the neighbors of the starting pixel level by level, which is more robust against stack overflow issues compared to a recursive DFS.
**Time:** `O(N * M)`, where `N` is the number of rows and `M` is the number of columns. Each pixel is enqueued and dequeued at most once. · **Space:** `O(N * M)` in the worst case. The space is used by the queue. The maximum size of the queue can be proportional to the number of pixels in the grid, for example, in a checkerboard-like pattern.
**Pros:** It is guaranteed to find the shortest path in terms of edges from the source (though not relevant for this problem).; It avoids the risk of `StackOverflowError` that can occur with deep recursion, making it more robust for large inputs.
**Cons:** The code can be slightly more verbose than the recursive DFS approach due to the explicit management of the queue.
### Explanation
Similar to the DFS approach, we first handle the edge case where the starting pixel's color is already the target color. We store the `initialColor` of the pixel at `(sr, sc)`. We initialize a queue (e.g., a `LinkedList` in Java) and add the starting coordinates `(sr, sc)` to it. We immediately change the color of the starting pixel `image[sr][sc]` to the new `color`. This is crucial to prevent adding the same pixel to the queue multiple times. We then enter a loop that continues as long as the queue is not empty. Inside the loop, we dequeue a pixel's coordinates. For this pixel, we examine its four neighbors (up, down, left, right). For each neighbor, we check if it's within the grid boundaries and if its color matches the `initialColor`. If a neighbor is valid (within bounds and has the correct color), we update its color to the new `color` and enqueue its coordinates for future processing. This process continues until the queue is empty, at which point all reachable pixels with the `initialColor` have been visited and recolored.

```java
import java.util.Queue;
import java.util.LinkedList;

class Solution {
    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int initialColor = image[sr][sc];
        if (initialColor == color) {
            return image;
        }

        int m = image.length;
        int n = image[0].length;
        Queue<int[]> queue = new LinkedList<>();
        
        queue.offer(new int[]{sr, sc});
        image[sr][sc] = color;

        int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

        while (!queue.isEmpty()) {
            int[] currentPixel = queue.poll();
            int r = currentPixel[0];
            int c = currentPixel[1];

            for (int[] dir : directions) {
                int newRow = r + dir[0];
                int newCol = c + dir[1];

                if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n && image[newRow][newCol] == initialColor) {
                    image[newRow][newCol] = color;
                    queue.offer(new int[]{newRow, newCol});
                }
            }
        }
        return image;
    }
}
```
### Algorithm
- Get the color of the starting pixel `(sr, sc)`, let's call it `initialColor`.
- If `initialColor` is the same as the target `color`, return the image.
- Create a queue and add the starting coordinates `(sr, sc)`.
- Update the starting pixel's color: `image[sr][sc] = color`.
- While the queue is not empty:
    - Dequeue a pixel `(r, c)`.
    - For each of its four neighbors `(nr, nc)`:
        - Check if the neighbor is within the grid boundaries.
        - Check if the neighbor's color `image[nr][nc]` is equal to `initialColor`.
        - If both checks pass, update the neighbor's color to `color` and enqueue the neighbor `(nr, nc)`.
- Return the modified `image`.

# Solutions
### Java

```java
class Solution {
private
  int[] dirs = {-1, 0, 1, 0, -1};
private
  int[][] image;
private
  int nc;
private
  int oc;
public
  int[][] floodFill(int[][] image, int sr, int sc, int color) {
    nc = color;
    oc = image[sr][sc];
    this.image = image;
    dfs(sr, sc);
    return image;
  }
private
  void dfs(int i, int j) {
    if (i < 0 || i >= image.length || j < 0 || j >= image[0].length ||
        image[i][j] != oc || image[i][j] == nc) {
      return;
    }
    image[i][j] = nc;
    for (int k = 0; k < 4; ++k) {
      dfs(i + dirs[k], j + dirs[k + 1]);
    }
  }
}

```

### CPP

```cpp
class Solution {
public:
  vector<vector<int>> floodFill(vector<vector<int>> &image, int sr, int sc,
                                int color) {
    int m = image.size(), n = image[0].size();
    int oc = image[sr][sc];
    int dirs[5] = {-1, 0, 1, 0, -1};
    function<void(int, int)> dfs = [&](int i, int j) {
      if (i < 0 || i >= m || j < 0 || j >= n || image[i][j] != oc ||
          image[i][j] == color) {
        return;
      }
      image[i][j] = color;
      for (int k = 0; k < 4; ++k) {
        dfs(i + dirs[k], j + dirs[k + 1]);
      }
    };
    dfs(sr, sc);
    return image;
  }
};

```

### Python

```python
class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]: def dfs(i, j): if (not 0 <= i < m or not 0 <= j < n or image[i][j] != oc or image[i][j] == color): return image[i][j] = color for a, b in pairwise(dirs): dfs(i + a, j + b) dirs = (- 1, 0, 1, 0, - 1) m, n = len(image), len(image[0]) oc = image[sr][sc] dfs(sr, sc) return image

```