Word Break II
<p>Given a string <code>s</code> and a dictionary of strings <code>wordDict</code>, add spaces in <code>s</code> to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in <strong>any order</strong>.</p>

<p><strong>Note</strong> that the same word in the dictionary may be reused multiple times in the segmentation.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;catsanddog&quot;, wordDict = [&quot;cat&quot;,&quot;cats&quot;,&quot;and&quot;,&quot;sand&quot;,&quot;dog&quot;]
<strong>Output:</strong> [&quot;cats and dog&quot;,&quot;cat sand dog&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;pineapplepenapple&quot;, wordDict = [&quot;apple&quot;,&quot;pen&quot;,&quot;applepen&quot;,&quot;pine&quot;,&quot;pineapple&quot;]
<strong>Output:</strong> [&quot;pine apple pen apple&quot;,&quot;pineapple pen apple&quot;,&quot;pine applepen apple&quot;]
<strong>Explanation:</strong> Note that you are allowed to reuse a dictionary word.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;catsandog&quot;, wordDict = [&quot;cats&quot;,&quot;dog&quot;,&quot;sand&quot;,&quot;and&quot;,&quot;cat&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 20</code></li>
	<li><code>1 &lt;= wordDict.length &lt;= 1000</code></li>
	<li><code>1 &lt;= wordDict[i].length &lt;= 10</code></li>
	<li><code>s</code> and <code>wordDict[i]</code> consist of only lowercase English letters.</li>
	<li>All the strings of <code>wordDict</code> are <strong>unique</strong>.</li>
	<li>Input is generated in a way that the length of the answer doesn&#39;t exceed&nbsp;10<sup>5</sup>.</li>
</ul>
Trie; Memoization; Array; Hash Table; String; Dynamic Programming; Backtracking
class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True

    def search(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                return False
            node = node.children[idx]
        return node.is_end


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        def dfs(s):
            if not s:
                return [[]]
            res = []
            for i in range(1, len(s) + 1):
                if trie.search(s[:i]):
                    for v in dfs(s[i:]):
                        res.append([s[:i]] + v)
            return res

        trie = Trie()
        for w in wordDict:
            trie.insert(w)
        ans = dfs(s)
        return [' '.join(v) for v in ans]
