<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:
    unordered_map<string,int> word_dict;
    int solve(string s, unordered_map<string, int> st,int start,vector<int> &dp)
    {
        int l=s.length();
        if(start == s.length())
        return 1;
        if(dp[start]!=-1)
        return dp[start];


        int i;
        int check ;
        string curr="";
        for(i=start;i<l;i++)
        {
            curr=curr+s[i];
            if(st.find(curr)!=st.end())
            {
                if(solve(s,st,i+1,dp))
                {
                    return dp[start]=1;
                }
            }
        }
        return dp[start]=0;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        word_dict.clear();
        vector<int> dp(301,-1);
        int i,j,n=wordDict.size();
        for(i=0;i<n;i++)
        {
            word_dict[wordDict[i]]=1;
        }

        if(solve(s,word_dict,0,dp)==1)
        return true;
        return false;
        
    }
};
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