<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 
{
    //vector<vector<string>> res;
    unordered_map<string,bool> memo;
public:
    bool wordBreak(const string& s, int start, set<string> &dict, vector<string>& words) 
    {
        if(start == s.size())
            return true;

        string newSentence = s.substr(start);
        if(memo.find(newSentence) != memo.end())
            return memo[newSentence];

        string word;
        for(int i=start; i<s.size(); i++)
        {
            word += s[i];
            if(dict.find(word) != dict.end())
            {
                // choose
                words.push_back(word);
                memo[word] = true;

                // explore
                string suffix = s.substr(i+1);
                bool ans = memo[suffix] = wordBreak(s, i+1, dict, words);
                if(ans)
                    return true;
                
                // un-choose
                words.pop_back();
            }
        }
        return memo[s];
    }
    
    bool wordBreak(string s, vector<string>& wordDict) 
    {
        memo.clear();
        set<string> dict{wordDict.begin(), wordDict.end()};
        vector<string> temp;
        return wordBreak(s,0,dict, temp);
    }
};
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