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 {
    Trie[] children = new Trie[26];
    boolean isEnd;

    void insert(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }

    boolean search(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                return false;
            }
            node = node.children[c];
        }
        return node.isEnd;
    }
}

class Solution {
    private Trie trie = new Trie();

    public List<String> wordBreak(String s, List<String> wordDict) {
        for (String w : wordDict) {
            trie.insert(w);
        }
        List<List<String>> res = dfs(s);
        return res.stream().map(e -> String.join(" ", e)).collect(Collectors.toList());
    }

    private List<List<String>> dfs(String s) {
        List<List<String>> res = new ArrayList<>();
        if ("".equals(s)) {
            res.add(new ArrayList<>());
            return res;
        }
        for (int i = 1; i <= s.length(); ++i) {
            if (trie.search(s.substring(0, i))) {
                for (List<String> v : dfs(s.substring(i))) {
                    v.add(0, s.substring(0, i));
                    res.add(v);
                }
            }
        }
        return res;
    }
}