+ problem 140

Author: fruit-in

Problemset/0140-Word Break II/README.md

@@ -0,0 +1,60 @@
+# 140. Word Break II
+Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
+
+**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
+
+#### Example 1:
+<pre>
+<strong>Input:</strong> s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
+<strong>Output:</strong> ["cats and dog","cat sand dog"]
+</pre>
+
+#### Example 2:
+<pre>
+<strong>Input:</strong> s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
+<strong>Output:</strong> ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
+<strong>Explanation:</strong> Note that you are allowed to reuse a dictionary word.
+</pre>
+
+#### Example 3:
+<pre>
+<strong>Input:</strong> s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
+<strong>Output:</strong> []
+</pre>
+
+#### Constraints:
+* `1 <= s.length <= 20`
+* `1 <= wordDict.length <= 1000`
+* `1 <= wordDict[i].length <= 10`
+* `s` and `wordDict[i]` consist of only lowercase English letters.
+* All the strings of `wordDict` are **unique**.
+* Input is generated in a way that the length of the answer doesn't exceed 10<sup>5</sup>.
+
+## Solutions (Python)
+
+### 1. Solution
+```Python
+from functools import cache
+
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
+        words = set(wordDict)
+
+        @cache
+        def backtracking(s: str) -> Optional[List[str]]:
+            ret = []
+
+            for i in range(1, min(len(s) + 1, 10)):
+                if s[:i] in wordDict:
+                    if i == len(s):
+                        return ret + [s]
+                    sentences = backtracking(s[i:])
+                    if sentences is not None:
+                        ret.extend("{} {}".format(s[:i], sentence)
+                                   for sentence in sentences)
+
+            return ret if ret != [] else None
+
+        return backtracking(s) if backtracking(s) is not None else []
+```

Problemset/0140-Word Break II/README_CN.md

@@ -0,0 +1,59 @@
+# 140. 单词拆分 II
+给定一个字符串 `s` 和一个字符串字典 `wordDict` ，在字符串 `s` 中增加空格来构建一个句子，使得句子中所有的单词都在词典中。**以任意顺序** 返回所有这些可能的句子。
+
+**注意：**词典中的同一个单词可能在分段中被重复使用多次。
+
+#### 示例 1:
+<pre>
+<strong>输入:</strong> s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
+<strong>输出:</strong> ["cats and dog","cat sand dog"]
+</pre>
+
+#### 示例 2:
+<pre>
+<strong>输入:</strong> s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
+<strong>输出:</strong> ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
+<strong>解释:</strong> 注意你可以重复使用字典中的单词。
+</pre>
+
+#### 示例 3:
+<pre>
+<strong>输入:</strong> s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
+<strong>输出:</strong> []
+</pre>
+
+#### 提示:
+* `1 <= s.length <= 20`
+* `1 <= wordDict.length <= 1000`
+* `1 <= wordDict[i].length <= 10`
+* `s` 和 `wordDict[i]` 仅有小写英文字母组成
+* `wordDict` 中所有字符串都 **不同**
+
+## 题解 (Python)
+
+### 1. 题解
+```Python
+from functools import cache
+
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
+        words = set(wordDict)
+
+        @cache
+        def backtracking(s: str) -> Optional[List[str]]:
+            ret = []
+
+            for i in range(1, min(len(s) + 1, 10)):
+                if s[:i] in wordDict:
+                    if i == len(s):
+                        return ret + [s]
+                    sentences = backtracking(s[i:])
+                    if sentences is not None:
+                        ret.extend("{} {}".format(s[:i], sentence)
+                                   for sentence in sentences)
+
+            return ret if ret != [] else None
+
+        return backtracking(s) if backtracking(s) is not None else []
+```

Problemset/0140-Word Break II/Solution.py

@@ -0,0 +1,23 @@
+from functools import cache
+
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
+        words = set(wordDict)
+
+        @cache
+        def backtracking(s: str) -> Optional[List[str]]:
+            ret = []
+
+            for i in range(1, min(len(s) + 1, 10)):
+                if s[:i] in wordDict:
+                    if i == len(s):
+                        return ret + [s]
+                    sentences = backtracking(s[i:])
+                    if sentences is not None:
+                        ret.extend("{} {}".format(s[:i], sentence)
+                                   for sentence in sentences)
+
+            return ret if ret != [] else None
+
+        return backtracking(s) if backtracking(s) is not None else []

README.md

@@ -128,6 +128,7 @@
 [137][137l]  |[Single Number II][137]                                                               |![rs]
 [138][138l]  |[Copy List with Random Pointer][138]                                                  |![py]
 [139][139l]  |[Word Break][139]                                                                     |![py]  ![rb]
+[140][140l]  |[Word Break II][140]                                                                  |![py]
 [141][141l]  |[Linked List Cycle][141]                                                              |![py]
 [142][142l]  |[Linked List Cycle II][142]                                                           |![py]
 [143][143l]  |[Reorder List][143]                                                                   |![py]
@@ -1723,6 +1724,7 @@
 [137]:Problemset/0137-Single%20Number%20II/README.md#137-single-number-ii
 [138]:Problemset/0138-Copy%20List%20with%20Random%20Pointer/README.md#138-copy-list-with-random-pointer
 [139]:Problemset/0139-Word%20Break/README.md#139-word-break
+[140]:Problemset/0140-Word%20Break%20II/README.md#140-word-break-ii
 [141]:Problemset/0141-Linked%20List%20Cycle/README.md#141-linked-list-cycle
 [142]:Problemset/0142-Linked%20List%20Cycle%20II/README.md#142-linked-list-cycle-ii
 [143]:Problemset/0143-Reorder%20List/README.md#143-reorder-list
@@ -3311,6 +3313,7 @@
 [137l]:https://leetcode.com/problems/single-number-ii/
 [138l]:https://leetcode.com/problems/copy-list-with-random-pointer/
 [139l]:https://leetcode.com/problems/word-break/
+[140l]:https://leetcode.com/problems/word-break-ii/
 [141l]:https://leetcode.com/problems/linked-list-cycle/
 [142l]:https://leetcode.com/problems/linked-list-cycle-ii/
 [143l]:https://leetcode.com/problems/reorder-list/

README_CN.md

@@ -128,6 +128,7 @@
 [137][137l]  |[只出现一次的数字 II][137]                                |![rs]
 [138][138l]  |[复制带随机指针的链表][138]                               |![py]
 [139][139l]  |[单词拆分][139]                                           |![py]  ![rb]
+[140][140l]  |[单词拆分 II][140]                                        |![py]
 [141][141l]  |[环形链表][141]                                           |![py]
 [142][142l]  |[环形链表 II][142]                                        |![py]
 [143][143l]  |[重排链表][143]                                           |![py]
@@ -1723,6 +1724,7 @@
 [137]:Problemset/0137-Single%20Number%20II/README_CN.md#137-只出现一次的数字-ii
 [138]:Problemset/0138-Copy%20List%20with%20Random%20Pointer/README_CN.md#138-复制带随机指针的链表
 [139]:Problemset/0139-Word%20Break/README_CN.md#139-单词拆分
+[140]:Problemset/0140-Word%20Break%20II/README_CN.md#140-单词拆分-ii
 [141]:Problemset/0141-Linked%20List%20Cycle/README_CN.md#141-环形链表
 [142]:Problemset/0142-Linked%20List%20Cycle%20II/README_CN.md#142-环形链表-ii
 [143]:Problemset/0143-Reorder%20List/README_CN.md#143-重排链表
@@ -3311,6 +3313,7 @@
 [137l]:https://leetcode.cn/problems/single-number-ii/
 [138l]:https://leetcode.cn/problems/copy-list-with-random-pointer/
 [139l]:https://leetcode.cn/problems/word-break/
+[140l]:https://leetcode.cn/problems/word-break-ii/
 [141l]:https://leetcode.cn/problems/linked-list-cycle/
 [142l]:https://leetcode.cn/problems/linked-list-cycle-ii/
 [143l]:https://leetcode.cn/problems/reorder-list/