vault backup: 2022-07-15 09:26:10

.obsidian/appearance.json

@@ -6,7 +6,5 @@
   "textFontFamily": "IBM Plex Mono,monospace",
   "monospaceFontFamily": "IBM Plex Mono",
   "baseFontSize": 19,
-  "enabledCssSnippets": [
-    "expander"
-  ]
+  "enabledCssSnippets": []
 }

.obsidian/plugins/mrj-text-expand/manifest.json

@@ -1,7 +1,7 @@
 {
   "id": "mrj-text-expand",
   "name": "Text expand",
-  "version": "0.10.8",
+  "version": "0.11.2",
   "description": "Search and paste/transclude links to located files.",
   "isDesktopOnly": false,
   "author": "MrJackphil",

OJ notes/pages/Leetcode Flood-Fill.md

@@ -5,7 +5,7 @@
 > ##### Algorithms:
 > #algorithm #BFS
 > ##### Data structures:
-> #DS #
+> #DS
 > ##### Difficulty:
 > #coding_problem #difficulty-easy
 > ##### Additional tags:
@@ -15,23 +15,118 @@
 
 ##### Related topics:
 ```expander
-tag:#<INSERT_TAG_HERE>
+tag:#BFS OR tag:#DFS
 ```
 
 
-
 ##### Links:
-- [Link to problem]()
+- [Link to problem](https://leetcode.com/problems/flood-fill/)
 ___
 ### 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:**
+
+![](https://assets.leetcode.com/uploads/2021/06/01/flood1-grid.jpg)
+
+```
+**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 #template_remove_me
+> 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
+
+```cpp
+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;
+  }
+};
+```