<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:
map<string,int> m;
vector<int> dp;

    bool solve(string s,int pl){
        if(s=="" || m[s]){
            return true;
        }
        // ) return 
        for(int i=1;i<=s.length();i++){
            // cout<<s.substr(0,i)<<" "<<dp[pl+i]<<" ";

            if(m[s.substr(0,i)]) {
                // cout<<s.substr(0,i)<<" "<<m[s.substr(0,i)];
                if(dp[pl+i]==-1) dp[pl+i]=solve(s.substr(i),i+pl);
                // cout<<s.substr(i)<<" "<<dp[i]<<"   ";
                if(dp[pl+i]==0) continue;
                else return true;
            }
        }
        return  false;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        for(string s:wordDict){
            m[s]=1;
            // cout<<s;
         }
         dp.resize(10000,-1);

         return solve(s,0);
    }
};
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