Tags: "leetcode", "recursion", "dfs", access_time 3-min read

Flood Fill

Created: May 11, 2020 by [lek-tin]

Last updated: May 11, 2020

An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).

Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, “flood fill” the image.

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 as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.

At the end, return the modified image.

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)), all pixels connected 
by a path of the same color as the starting pixel 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.

Note:

• The length of image and image[0] will be in the range [1, 50].
• The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
• The value of each color in image[i][j] and newColor will be an integer in [0, 65535].

Solution (DFS using a queue)

Java

Solution (DFS using recursion)

Java

class Solution {
    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        if (image.length < 0 || image[0].length < 0 || sr < 0 || sr >= image.length || sc < 0 || sc >= image[0].length) {
            return image;
        }

        dfs(image, sr, sc, image[sr][sc], newColor);

        // recover the real colors by flipping the sign
        for (int i = 0; i < image.length; i++) {
            for (int j = 0; j < image[0].length; j++) {
                if (image[i][j] < 0) {
                    image[i][j] = -image[i][j];
                }
            }
        }

        return image;
    }

    private void dfs(int[][] image, int row, int col, int oldColor, int newColor) {
        if (image.length < 0 || image[0].length < 0 || row < 0 || row >= image.length || col < 0 || col >= image[0].length || image[row][col] < 0) {
            return;
        }

        // -newColor means this cell was visited
        image[row][col] = -newColor;

        int[][] dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};
        for (int[] dir: dirs) {
            int new_r = row + dir[0];
            int new_c = col + dir[1];
            if ( new_r >=0 && new_r < image.length && new_c >= 0 && new_c < image[0].length && image[new_r][new_c] == oldColor) {
                dfs(image, new_r, new_c, oldColor, newColor);
            }
        }
    }
}