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.
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.
Pseudocode
starting from an initial point in the matrix
find neighbouring cells that are of value 1, if it isn't 1, then return/continue
if it is, then reassign value to color
to traverse matrix, either dfs or bfs
dfs - recurse over 4 directions
bfs - push cells with value 1 into queue and iterate over 4 directions
Solution
// Some code
var floodFill = function (image, sr, sc, color) {
const targetColor = image[sr][sc];
let mLength = image.length;
let nLength = image[sr].length;
function walk(imgArr, m, n) {
// base condition
if (m < 0 || n < 0 || m >= mLength || n >= nLength) {
return;
}
if (imgArr[m][n] !== 1) {
return;
}
if (imgArr[m][n] === targetColor) {
imgArr[m][n] = color;
} else {
return;
}
// pre
// recurse
const walkLeft = walk(imgArr, m + 1, n);
const walkDown = walk(imgArr, m, n + 1);
const walkRight = walk(imgArr, m - 1, n);
const walkUp = walk(imgArr, m, n - 1);
// post
return;
}
walk(image, sr, sc);
return image;
};
Time and Space Complexity
Time
dfs - time complexity of recursion is O(M*N)
bfs - similar time complexity O(M*N)
Total - O(M*N)
Space
dfs - space requirement for recursive function is O(M*N)
bfs - space requirement for bfs queue is O(M + N) ? not every cell is pushed into queue
