LeetCode 733. Flood Fill

Description

https://leetcode.com/problems/flood-fill/

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 srsc, and newColor. 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 newColor.

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, newColor = 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, newColor = 2
Output: [[2,2,2],[2,2,2]]

Constraints:

  • m == image.length
  • n == image[i].length
  • 1 <= m, n <= 50
  • 0 <= image[i][j], newColor < 216
  • 0 <= sr < m
  • 0 <= sc < n

Explanation

Starting from image position[sr][sc] to search all the positions with the same old color and then replace with the new color.

Python Solution

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, newColor: int) -> List[List[int]]:
        
        old_color = image[sr][sc]
        
        if old_color != newColor:
            self.helper(image, sr, sc, old_color, newColor)
        
        return image
        
    def helper(self, image, i, j, old_color, new_color):
        
        if i < 0 or i > len(image) - 1:
            return
        
        if j < 0 or j > len(image[0]) - 1:
            return
        
        if image[i][j] != old_color:
            return
        else:
            image[i][j] = new_color
            
            self.helper(image, i, j - 1, old_color, new_color)            
            self.helper(image, i, j + 1, old_color, new_color)            
            self.helper(image, i - 1, j, old_color, new_color)            
            self.helper(image, i + 1, j, old_color, new_color)
        
  • Time Complexity: O(MN).
  • Space Complexity: O(1).

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