chore: add daily leetcode post 994.腐烂的橘子_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/994.腐烂的橘子_translated.md

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+---
+title: 994.Rotten orange.md
+date: 2024.05.14 0:00
+tags:
+  - Python
+  - BFS
+  - Bilateral queue
+abbrlink: 56e64fdd
+---
+
+# topic：
+
+"""
+<p>Given&nbsp;<code>m x n</code>&nbsp;grid
+ <meta charset="UTF-8" />&nbsp;<code>grid</code>&nbsp;middle，Each cell can have one of the following three values：</p>
+
+<ul> 
+ <li>value&nbsp;<code>0</code>&nbsp;Represents the empty unit；</li> 
+ <li>value&nbsp;<code>1</code>&nbsp;Represents fresh oranges；</li> 
+ <li>value&nbsp;<code>2</code>&nbsp;代表Rotten orange。</li> 
+</ul>
+
+<p>every minute，Rotten orange&nbsp;<strong>around&nbsp;4 Adjacent in this direction</strong> Fresh oranges will rot。</p>
+
+<p>return <em>直到单元格middle没有新鲜橘子为止所必须经过的最小分钟数。If it is impossible，return&nbsp;<code>-1</code></em>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>Exemplary example 1：</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/02/16/oranges.png" style="height: 137px; width: 650px;" /></strong></p>
+
+<pre>
+<strong>enter：</strong>grid = [[2,1,1],[1,1,0],[0,1,1]]
+<strong>Output：</strong>4
+</pre>
+
+<p><strong>Exemplary example 2：</strong></p>
+
+<pre>
+<strong>enter：</strong>grid = [[2,1,1],[0,1,1],[1,0,1]]
+<strong>Output：</strong>-1
+<strong>explain：</strong>Orange in the lower left corner（First 2 OK， First 0 List）Never rot，Because rotten will only happen in 4 In the direction。
+</pre>
+
+<p><strong>Exemplary example 3：</strong></p>
+
+<pre>
+<strong>enter：</strong>grid = [[0,2]]
+<strong>Output：</strong>0
+<strong>explain：</strong>because 0 There is no fresh orange in minutes，So the answer is 0 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>hint：</strong></p>
+
+<ul> 
+ <li><code>m == grid.length</code></li> 
+ <li><code>n == grid[i].length</code></li> 
+ <li><code>1 &lt;= m, n &lt;= 10</code></li> 
+ <li><code>grid[i][j]</code> Only for&nbsp;<code>0</code>、<code>1</code>&nbsp;or&nbsp;<code>2</code></li> 
+</ul>
+
+<div><div>Related Topics</div><div><li>Priority search</li><li>Array</li><li>matrix</li></div></div><br><div><li>👍 872</li><li>👎 0</li></div>
+"""
+
+# Thought：
+
+这个问题可以用Priority search（BFS）To solve。We need to track the spread of rotten oranges，Record time，And check if there is a fresh orange that cannot be rotten。The initial idea of ​​the original idea：
+```python
+class Solution:
+    def orangesRotting(self, grid: List[List[int]]) -> int:
+        bad_orange = []
+        # 找到所有初始Rotten orange
+        for i in range(len(grid)):
+            for j in range(len(grid[0])):
+                if grid[i][j] == 2:
+                    # 存入初始队List
+                    bad_orange.append((i, j))
+```
+Similar to multi -threaded，每个线程存入一个初始队List，初始队List通过BFSGradual diffusion
+
+# Code：
+
+```python
+from collections import deque
+
+class Solution:
+    def orangesRotting(self, grid: List[List[int]]) -> int:
+        bad_orange = deque()
+        fresh_oranges = 0
+        rows, cols = len(grid), len(grid[0])
+        
+        # 找到所有初始Rotten orange，And calculate the number of fresh oranges
+        for i in range(rows):
+            for j in range(cols):
+                if grid[i][j] == 2:
+                    bad_orange.append((i, j))
+                elif grid[i][j] == 1:
+                    fresh_oranges += 1
+        
+        # 方向Array：up down left right
+        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
+        
+        # If there is no fresh orange，直接return 0
+        if fresh_oranges == 0:
+            return 0
+        
+        # BFS
+        minutes = 0
+        while bad_orange:
+            minutes += 1
+            for _ in range(len(bad_orange)):
+                x, y = bad_orange.popleft()
+                for dx, dy in directions:
+                    nx, ny = x + dx, y + dy
+                    if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] == 1:
+                        grid[nx][ny] = 2
+                        fresh_oranges -= 1
+                        bad_orange.append((nx, ny))
+        
+        # If there are fresh oranges，return -1
+        return minutes - 1 if fresh_oranges == 0 else -1
+```
+
+```go
+import (
+	"sync"
+)
+
+func orangesRotting(grid [][]int) int {
+	rows, cols := len(grid), len(grid[0])
+	badOranges := make([][2]int, 0)
+	freshOranges := 0
+
+	// 找到所有初始Rotten orange，And calculate the number of fresh oranges
+	for r := 0; r < rows; r++ {
+		for c := 0; c < cols; c++ {
+			if grid[r][c] == 2 {
+				badOranges = append(badOranges, [2]int{r, c})
+			} else if grid[r][c] == 1 {
+				freshOranges += 1
+			}
+		}
+	}
+
+	// If there is no fresh orange，直接return 0
+	if freshOranges == 0 {
+		return 0
+	}
+
+	directions := [][2]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}}
+	minutes := 0
+
+	var wg sync.WaitGroup
+
+	// BFS
+	for len(badOranges) > 0 {
+		minutes++
+		nextBadOranges := make([][2]int, 0)
+		for _, orange := range badOranges {
+			x, y := orange[0], orange[1]
+			wg.Add(1)
+			go func(x, y int) {
+				defer wg.Done()
+				for _, d := range directions {
+					nx, ny := x+d[0], y+d[1]
+					if nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == 1 {
+						grid[nx][ny] = 2
+						nextBadOranges = append(nextBadOranges, [2]int{nx, ny})
+						freshOranges--
+					}
+				}
+			}(x, y)
+		}
+		wg.Wait()
+		badOranges = nextBadOranges
+	}
+
+	// If there are fresh oranges，return -1
+	if freshOranges > 0 {
+		return -1
+	}
+	return minutes - 1
+}
+```