Flood Fill Algorithm

Moderate
0/80
Average time to solve is 10m
45 upvotes
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Problem statement

Ninja has found his new passion for photography. He has clicked some really good photos but in one of his images, he doesn’t like the color of a particular region. So, he decides to change the color of that region. Can you help him with this task?

The image is represented in the form of a 2D array of size M * N. Each pixel in the image is a positive integer. Ninja has given you the coordinates (row and column) of a certain pixel and the new color value. You need to replace the color of the given pixel and all adjacent same-colored pixels with the new color.

Note: 
Two pixels are adjacent if they are connected to each other in any of the four directions: up, down, left, or right.

Diagonal pixels are not considered adjacent.
Example: 
Consider the image of size 4*4, shown below (left). Let the coordinates of the starting pixel are (1, 2) and the new color is 8. The starting pixel, highlighted with red color, has a pixel value of 3. 

On replacing the given pixel and all adjacent same-colored pixels with the new color we get the new image, shown below (right). The modified pixels are highlighted with green color.

example

Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of input contains an integer ‘T’ representing the number of test cases.

The first line of every test case contains two space-separated integers ‘M’ and ‘N’ representing the number of rows and columns in the image.

Each of the next ‘M’ lines contains ‘N’ space-separated integers denoting the pixel values of the image.

    The next line contains three space-separated integers ‘X’, ‘Y’, and ‘C’ denoting the row and column of the starting pixel and the new color, respectively.
Output Format: 
For each test case, the newly colored image is printed in the form of an M * N Matrix.
Note: 
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 100 
1 <= M, N <= 50
0 <= X < M
0 <= Y < N
1 <= Image[i][j], C <= 10^5    

Time Limit: 1 sec
Sample Input 1:
2
4 4
1 1 7 5
1 3 3 3
6 5 5 3
2 2 3 3
1 2 8
5 6
1 2 1 2 3 5
1 2 2 4 3 4
1 2 4 4 5 4
6 2 2 2 3 4
7 6 1 3 3 3
3 2 4
Sample Output 1:
1 1 7 5
1 8 8 8
6 5 5 8
2 2 8 8
1 4 1 2 3 5
1 4 4 4 3 4
1 4 4 4 5 4
6 4 4 4 3 4
7 6 1 3 3 3
Explanation For Sample Input 1:
For the first test case, refer to the example explained before.

For the second test case, we are given an image of size 5*6, shown below (left). The coordinates of the starting pixel are (3, 2) and the new color is 4. The starting pixel, highlighted with red color, has a pixel value of 2. 

On replacing the given pixel and all adjacent same colored pixels with the new color we get the new image, shown below (right). The modified pixels are highlighted with green color.

Sample 1 - TestCase 2

Sample Input 2:
2
1 1
5
0 0 6
2 2
1 3
3 3
1 1 5
Sample Output 2:
6
1 5
5 5
Hint

Can you solve the problem using recursion?

Approaches (2)
Using Recursion
  • Starting from the given pixel let's traverse the image using recursion.
  • The idea is to replace the colour of the starting pixel with the new colour and repeat the same procedure for the adjacent same coloured pixels.
  • During the traversal, we do not replace the colour of the current pixel if:
    • It has a different colour than the starting pixel.
    • Or if it has already been replaced with a new colour.
  • In this way, we generate a new image.
  • This approach is the same as applying DFS on the given image.

 

Algorithm:

 

  • Start with the given pixel.
  • Base Condition 1: If the pixel has a different color than the starting pixel, return.
  • Base Condition 2: If the pixel has already been replaced with the new color, return.
  • Replace the colour of the current pixel with the new colour.
  • Recur for the adjacent pixels (up, down, left and right).
  • Return the new image.
Time Complexity

O(M * N) per test case, where M and N are the number of rows and columns in the image, respectively.

 

In the worst case, we will be traversing the complete image. Hence, the overall time complexity is O(M * N)

Space Complexity

O(M * N) per test case, where M and N are the number of rows and columns in the image, respectively.

 

In the worst case, O(M * N) extra space is required by the recursion stack. Hence the overall space complexity is O(M * N).

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