Flood Fill

Problem 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.

Examples

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: 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, color = 0
Output: [[0,0,0],[0,0,0]]
Explanation: The starting pixel is already colored 0, so no changes are made to the image.

Constraints

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

Solution for Flood Fill

Approach: Depth-First Search

Intuition

We perform the algorithm explained in the problem description: paint the starting pixels, plus adjacent pixels of the same color, and so on.

Algorithm

Say color is the color of the starting pixel. Let's flood fill the starting pixel: we change the color of that pixel to the new color, then check the 4 neighboring pixels to make sure they are valid pixels of the same color, and of the valid ones, we flood fill those, and so on.

We can use a function dfs to perform a flood fill on a target pixel.

Code in Different Languages

C++

Written by @Shreyash3087

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

Java

Written by @Shreyash3087

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

Python

Written by @Shreyash3087

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        R, C = len(image), len(image[0])
        color = image[sr][sc]
        if color == newColor:
            return image
        def dfs(r, c):
            if image[r][c] == color:
                image[r][c] = newColor
                if r >= 1:
                    dfs(r-1, c)
                if r + 1 < R:
                    dfs(r + 1, c)
                if c >= 1:
                    dfs(r, c - 1)
                if c + 1 < C:
                    dfs(r, c + 1)

        dfs(sr, sc)
        return image

Complexity Analysis

Time Complexity: $O(N)$

Reason: where N is the number of pixels in the image. We might process every pixel.

Space Complexity: $O(N)$

Reason: the size of the implicit call stack when calling dfs.

References

• LeetCode Problem: Flood Fill

• Solution Link: Flood Fill

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