<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 {
    int n;
    vector<int> a;
    unordered_set<string> w;
public:
    
    bool recursive_check(string& x, int start){
        if(a[start]==0){return false;}
        if(a[start]==1){return true;}
        
        string curr = x.substr(start);
        if(w.find(curr)!=w.end()){
            a[start]=1;
            return true;
        }
        
        //3,4
        for(int i=n-1;i>start;i--){
            curr.pop_back();
            if(w.find(curr)!=w.end()){
                bool flag = recursive_check(x,i);
                if(flag){
                    a[start]=1;
                    return true;
                }
            }
        }
        a[start]=0;
        return false;
    }
    
    bool wordBreak(string s, vector<string>& wordDict) {
        for(auto it : wordDict){
            w.insert(it);
        }
        n=s.length();
        a.assign(n,-1);
        
        //for(int i=n-1;i>=0;i--){
         //   recursive_check(s,i);
        //}
        recursive_check(s,0);
        if(a[0]==1){
            return true;
        }
        
        return false;
    }
};
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