733. Flood Fill
https://leetcode.com/problems/flood-fill/
The length of
The given starting pixel will satisfy
The value of each color in
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DFS - recursive - faster than BFS
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:
image
and image[0]
will be in the range [1, 50]
.0 <= sr < image.length
and 0 <= sc < image[0].length
.image[i][j]
and newColor
will be an integer in [0, 65535]
.
Note - check for newColor == origColor corner case to avoid infinite loop
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Related problems
island-perimeter
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Related problems
island-perimeter
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BFS - Non recursive - Queue based
Time - O(r * c)
Space - O(r * c)
Space - O(r * c)
DFS - recursive - faster than BFS
Time - O(r * c)