notes/OJ notes/pages/Leetcode Flood-Fill.md

Leetcode Flood-Fill

2022-07-15 09:01

Algorithms:

#algorithm #BFS

Data structures:

#DS

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

---

Problem

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 <= m, n <= 50
• 0 <= image[i][j], color < 216
• 0 <= sr < m
• 0 <= sc < n

Thoughts

[!summary] This is a search problem, can be solved using DFS or BFS

This one can be optimized.

Initially, I wanted to use a hash map to record cells that are visited, but the this takes up extra space.

Then I found out that I can use colors:

• if image[r][c] == color, this means
  • the cell is no need to correct(color == origcolor),
  • or this has been corrected so I don't have to go over again.

There are checks in the loop:

• check the color is not visited, as shown above
• check the coord is not OOB
• check the cell is equal to origColor, to only fill same origColor.

#TODO: Write in DFS

Solution

class Solution {
public:
  vector<vector<int>> floodFill(vector<vector<int>> &image, int sr, int sc,
                                int color) {
    int origColor = image[sr][sc];
    queue<pair<int, int>> todo;
    int m = image.size();
    int n = image[0].size();

    todo.push({sr, sc});
    int r, c;

    while (!todo.empty()) {
      r = todo.front().first;
      c = todo.front().second;
      todo.pop();

      if (image[r][c] != origColor || image[r][c] == color) {
        // already colored
        continue;
      }

      image[r][c] = color;

      if (r > 0) {
        todo.push({r - 1, c});
      }
      if (r < m - 1) {
        todo.push({r + 1, c});
      }
      if (c > 0) {
        todo.push({r, c - 1});
      }
      if (c < n - 1) {
        todo.push({r, c + 1});
      }
    }

    return image;
  }
};