733. Flood Fill - Explanation

Problem Link

Description

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:

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 == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• 0 <= image[i][j], color < (2^16)
• 0 <= sr < m
• 0 <= sc < n

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1. Depth First Search

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]:
        if image[sr][sc] == color:
            return image

        m, n = len(image), len(image[0])
        dirs = [1,0,-1,0,1]

        def dfs(r, c, org):
            if not (0 <= r < m) or not (0 <= c < n) or image[r][c] != org:
                return

            image[r][c] = color
            for d in range(4):
                dfs(r + dirs[d], c + dirs[d + 1], org)

        dfs(sr, sc, image[sr][sc])
        return image

Time & Space Complexity

• Time complexity: $O(m * n)$
• Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the image.

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2. Breadth First Search

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]:
        orig = image[sr][sc]
        if orig == color:
            return image

        m, n = len(image), len(image[0])
        q = deque([(sr, sc)])
        image[sr][sc] = color
        dirs = [(1,0), (-1,0), (0,1), (0,-1)]

        while q:
            r, c = q.popleft()
            for dr, dc in dirs:
                nr, nc = r + dr, c + dc
                if 0 <= nr < m and 0 <= nc < n and image[nr][nc] == orig:
                    image[nr][nc] = color
                    q.append((nr, nc))

        return image

Time & Space Complexity

• Time complexity: $O(m * n)$
• Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the image.