Flood fill is an algorithm mainly used to determine a bounded area connected to a given node in a multi-dimensional array. It is a close resemblance to the bucket tool in paint programs.

The most approached implementation of the algorithm is a stack-based recursive function, and that’s what we’re gonna talk about next.

**How does it work?**

The problem is pretty simple and usually follows these steps:

- Take the position of the starting point.
- Decide wether you want to go in 4 directions (
**N, S, W, E**) or 8 directions (**N, S, W, E, NW, NE, SW, SE**). - Choose a replacement color and a target color.
- Travel in those directions.
- If the tile you land on is a target, reaplce it with the chosen color.
- Repeat 4 and 5 until you’ve been everywhere within the boundaries.

Let’s take the following array as an example:

The red square is the starting point and the gray squares are the so called walls.

For further details, here’s a piece of code describing the function:

```
int wall = -1;
void flood_fill(int pos_x, int pos_y, int target_color, int color)
{
if(a[pos_x][pos_y] == wall || a[pos_x][pos_y] == color) // if there is no wall or if i haven't been there
return; // already go back
if(a[pos_x][pos_y] != target_color) // if it's not color go back
return;
a[pos_x][pos_y] = color; // mark the point so that I know if I passed through it.
flood_fill(pos_x + 1, pos_y, color); // then i can either go south
flood_fill(pos_x - 1, pos_y, color); // or north
flood_fill(pos_x, pos_y + 1, color); // or east
flood_fill(pos_x, pos_y - 1, color); // or west
return;
}
```

As seen above, my starting point is (4,4). After calling the function for the start coordinates **x = 4** and **y = 4**, I can start checking if there is no wall or color on the spot. If that is valid i mark the spot with one **“color”** and start checking the other adiacent squares.

Going south we will get to point (5,4) and the function runs again.

**Excercise problem**

I always considered that solving a (or more) problem/s using a newly learned algorithm is the best way to fully understand the concept.

So here’s one:

**Statement:**

In a bidimensional array you are given n number of **“islands”**. Try to find the largest area of an island and the corresponding island number. 0 marks water and any other x between 1 and n marks one square from the surface corresponding to island x.

**Input**

**n**- the number of islands.**l,c**- the dimensions of the matrix.- the next
**l**lines,**c**numbers giving the**l**th row of the matrix.

**Output**

**i**- the number of the island with the largest area.**A**- the area of the**i**‘th island.

**Ex:**

You have the following input:

```
2 4 4
0 0 0 1
0 0 1 1
0 0 0 2
2 2 2 2
```

For which you will get island no. 2 as the biggest island with the area of 5 squares.

**Hints**

The problem is quite easy, but here are some hints:

```
1. Use the flood-fill algorithm whenever you encounter a new island.
2. As opposed to the sample code, you should go through the area of the island and not on the ocean (0 tiles).
```