Flood Fill Python Solution with Tests

Tested Python solution for LeetCode 733 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 733, Easy. Topics: Array, Depth-First Search, Breadth-First Search, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 12 parametrized pytest cases, and a playground notebook:

lcpy gen -n 733   # by problem number
lcpy gen -s flood_fill   # by problem name

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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.

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Examples

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 horizontally or vertically connected to the starting pixel.

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.

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Constraints

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

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m*n)
    # Space: O(m*n)
    def flood_fill(self, image: list[list[int]], sr: int, sc: int, color: int) -> list[list[int]]:
        original = image[sr][sc]
        if original == color:
            return image

        def dfs(r: int, c: int) -> None:
            if r < 0 or r >= len(image) or c < 0 or c >= len(image[0]) or image[r][c] != original:
                return
            image[r][c] = color
            for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
                dfs(r + dr, c + dc)

        dfs(sr, sc)
        return image

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Complexity

Time	Space
O(m*n)	O(m*n)

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Tags

Grind 75, Grind, NeetCode All.