0733. Flood Fill

0733. Flood Fill#

Related Problem:

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:

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

Analysis#

  • DFS:

Solution#

"""A solution using DFS"""
def solution(image, sr, sc, color):

    m, n = len(image), len(image[0])

    orig_color = image[sr][sc]

    # special case:
    # special case has to be considered here
    # if the condition is not met, the dfs will run forever
    # therefore, we have to break if the condition is met
    if orig_color == color:
        return image

    def dfs(image, r, c, color):
        if image[r][c] == orig_color:
            image[r][c] = color 
            # dfs on its neighbors
            # left neighbor
            if r > 0:
                dfs(image, r - 1, c, color)
            # right neighbor
            if r < m - 1:
                dfs(image, r + 1, c, color)
            # up neighbor
            if c > 0:
                dfs(image, r, c - 1, color)
            # down neighbor
            if c < n - 1:
                dfs(image, r, c + 1, color)

    dfs(image, sr, sc, color)

    return image


# tests
image = [[0,0,0],[0,0,0]]
sr = 0
sc = 0
color = 0
print(solution(image, sr, sc, color))

image = [[1,1,1],[1,1,0],[1,0,1]]
sr = 1
sc = 1
color = 2
print(solution(image, sr, sc, color))

[[0, 0, 0], [0, 0, 0]]
[[2, 2, 2], [2, 2, 0], [2, 0, 1]]

"""A breadth first search - non recursive implementation"""
def solution(image, sr, sc, color):

    m, n = len(image), len(image[0])

    orig_color = image[sr][sc]

    # initialize a queue storing all nodes in sequence generated by BFS
    adjs = [(sr, sc)]
    
    # special case:
    # special case has to be considered here
    # if the condition is not met, this will run foever
    if orig_color == color:
        return image

    while adjs:
        r, c = adjs.pop(0)

        # fill in color if condition met
        if image[r][c] == orig_color:
            image[r][c] = color 
        
            # append adjacent nodes to the queue
            # left
            if r > 0:
                adjs.append((r - 1, c))
            # right
            if r < m - 1:
                adjs.append((r + 1, c))
            # up
            if c > 0:
                adjs.append((r, c - 1))
            # down
            if c < n - 1:
                adjs.append((r, c + 1))
        
    return image

# tests
image = [[0,0,0],[0,0,0]]
sr = 0
sc = 0
color = 0
print(solution(image, sr, sc, color))

image = [[1,1,1],[1,1,0],[1,0,1]]
sr = 1
sc = 1
color = 2
print(solution(image, sr, sc, color))            

[[0, 0, 0], [0, 0, 0]]
[[2, 2, 2], [2, 2, 0], [2, 0, 1]]
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