<p>Given a string <code>s</code> and a dictionary of strings <code>wordDict</code>, return <code>true</code> if <code>s</code> can be segmented into a space-separated sequence of one or more dictionary words.</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;leetcode&quot;, wordDict = [&quot;leet&quot;,&quot;code&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> Return true because &quot;leetcode&quot; can be segmented as &quot;leet code&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;applepenapple&quot;, wordDict = [&quot;apple&quot;,&quot;pen&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> Return true because &quot;applepenapple&quot; can be segmented as &quot;apple pen apple&quot;.
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> false
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 300</code></li>
	<li><code>1 &lt;= wordDict.length &lt;= 1000</code></li>
	<li><code>1 &lt;= wordDict[i].length &lt;= 20</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>
</ul>

class Solution {
 public:
  bool wordBreak(string s, vector<string>& wordDict) {
    const int n = s.length();
    const int maxLength = getMaxLength(wordDict);
    const unordered_set<string> wordSet{wordDict.begin(), wordDict.end()};
    // dp[i] := true if s[0..i) can be segmented
    vector<int> dp(n + 1);
    dp[0] = true;

    for (int i = 1; i <= n; ++i)
      for (int j = i - 1; j >= 0; --j) {
        if (i - j > maxLength)
          break;
        if (dp[j] && wordSet.contains(s.substr(j, i - j))) {
          dp[i] = true;
          break;
        }
      }

    return dp[n];
  }

 private:
  int getMaxLength(const vector<string>& wordDict) {
    return ranges::max_element(wordDict, ranges::less{},
                               [](const string& word) {
      return word.length();
    })->length();
  }
};
10000