Flood Fill

Description:​

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.

Video Explanation:​

Approaches:​

1. Depth-First Search (DFS) (Optimal)​

This is a classic graph traversal problem on a 2D matrix. We want to start at the given pixel (sr, sc) and change its color to the new color. Then, we recursively visit all of its 4-directional neighbors (up, down, left, right) that share the exact same original color as the starting pixel, and update their colors as well.

Algorithm:

1. Check if the starting pixel image[sr][sc] is already the new color. If it is, return the image immediately to avoid an infinite loop.
2. Store the original color of the starting pixel in a variable initialColor.
3. Launch a DFS helper function from (sr, sc).
4. In the DFS function:
   • Check boundaries: If the current row or column is out of bounds, return.
   • Check color: If the current pixel's color does not match initialColor, return.
   • Update the pixel's color to the new color.
   • Recursively call the DFS function for the pixel's 4 neighbors: (r-1, c), (r+1, c), (r, c-1), (r, c+1).
5. Return the modified image.

Complexity​

• Time Complexity: $O(M \times N)$ where $M$ is the number of rows and $N$ is the number of columns. In the worst-case scenario, we might have to process and recolor every single pixel in the image.
• Space Complexity: $O(M \times N)$ for the recursion call stack in the worst case (e.g., if the entire image is the same color and the recursion goes deep).

Solutions:​

C++

class Solution {
private:
    void dfs(vector<vector<int>>& image, int r, int c, int initialColor, int newColor) {
        // Boundary checks and color check
        if (r < 0 || r >= image.size() || c < 0 || c >= image[0].size() || image[r][c] != initialColor) {
            return;
        }
        
        // Update color
        image[r][c] = newColor;
        
        // 4-directional DFS
        dfs(image, r - 1, c, initialColor, newColor); // Up
        dfs(image, r + 1, c, initialColor, newColor); // Down
        dfs(image, r, c - 1, initialColor, newColor); // Left
        dfs(image, r, c + 1, initialColor, newColor); // Right
    }
    
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        int initialColor = image[sr][sc];
        
        // Avoid infinite loop if the color is already the target color
        if (initialColor != color) {
            dfs(image, sr, sc, initialColor, color);
        }
        
        return image;
    }
};

Java

class Solution {
    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int initialColor = image[sr][sc];
        
        if (initialColor != color) {
            dfs(image, sr, sc, initialColor, color);
        }
        
        return image;
    }
    
    private void dfs(int[][] image, int r, int c, int initialColor, int newColor) {
        if (r < 0 || r >= image.length || c < 0 || c >= image[0].length || image[r][c] != initialColor) {
            return;
        }
        
        image[r][c] = newColor;
        
        dfs(image, r - 1, c, initialColor, newColor);
        dfs(image, r + 1, c, initialColor, newColor);
        dfs(image, r, c - 1, initialColor, newColor);
        dfs(image, r, c + 1, initialColor, newColor);
    }
}

Python

class Solution:
    def floodFill(self, image: list[list[int]], sr: int, sc: int, color: int) -> list[list[int]]:
        initialColor = image[sr][sc]
        
        if initialColor == color:
            return image
            
        def dfs(r, c):
            if r < 0 or r >= len(image) or c < 0 or c >= len(image[0]) or image[r][c] != initialColor:
                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

JavaScript

/**
 * @param {number[][]} image
 * @param {number} sr
 * @param {number} sc
 * @param {number} color
 * @return {number[][]}
 */
var floodFill = function(image, sr, sc, color) {
    const initialColor = image[sr][sc];
    
    if (initialColor === color) return image;
    
    const dfs = (r, c) => {
        if (r < 0 || r >= image.length || c < 0 || c >= image[0].length || image[r][c] !== initialColor) {
            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;
};