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Solution 15: Chromatic Saturation (Flood Fill)

Approach Explanation

We use DFS to visit all connected pixels of the same original color and change them to the new color.

Step-by-Step Logic

  1. Store the start_color = image[sr][sc].
  2. If start_color == color, return image.
  3. Implement dfs(r, c):
    • If (r, c) is out of bounds or image[r][c] != start_color, return.
    • Set image[r][c] = color.
    • Call dfs on 4 neighbors.
  4. Call dfs(sr, sc) and return image.

Complexity

  • Time Complexity: O(M * N).
  • Space Complexity: O(M * N).

Code

def flood_fill(image, sr, sc, color):
    start_color = image[sr][sc]
    if start_color == color: return image
    
    rows, cols = len(image), len(image[0])
    
    def dfs(r, c):
        if (r < 0 or r >= rows or 
            c < 0 or c >= cols or 
            image[r][c] != start_color):
            return
            
        image[r][c] = color
        dfs(r + 1, c)
        dfs(r - 1, c)
        dfs(r, c + 1)
        dfs(r, c - 1)
        
    dfs(sr, sc)
    return image