An image is represented by an
m x n integer grid
image[i][j] represents the pixel value of the image.
You are also given three integers
newColor. You should perform a flood fill on the image starting from the pixel
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
Return the modified image after performing the flood fill.
Input: image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, newColor = 2
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.
Input: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, newColor = 2
m == image.length
n == image[i].length
1 <= m, n <= 50
0 <= image[i][j], newColor < 216
0 <= sr < m
0 <= sc < n
We can move in at most four directions from each cell, i.e., up, down, left, and right. To move to another cell, the neighboring cell must have a value equal to the current cell’s value. In this case, we’ll simply call DFS on the provided cell, and it will recursively assign the new value to its neighbors. We also need to make sure that we don’t process a cell whose value has already been changed.
Let’s look at the code for the solution:
The time complexity will be O(n), where n is the number of cells in the array. We might go through every cell.
The space complexity will be O(n), which is, the size of the stack as we recursively call DFS.