<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,bool> dict;
    string str;
    int n;
    int dp[301][301];
    bool rec(int prev, int curr)
    {
        string s = str.substr(prev,curr-prev+1);
        bool ans = 0;
        if(curr == n)
        {
            if(prev == n)
            {
                return 1;
            }
            else
            {
                return 0;
            }
        }

        // cache check
        if(dp[prev][curr] != -1)
        {
            return dp[prev][curr];
        }

        if(dict[s])
        {
         ans |= rec(curr+1,curr+1);   
         ans |= rec(prev,curr+1);
        }
        else
        {
            ans |= rec(prev,curr+1);
        }
        dp[prev][curr] = ans;
        return ans;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        for(auto x: wordDict)
        {
            dict[x] = 1;
        }
        str = s;
        n = s.length();
        memset(dp,-1,sizeof(dp));
        return rec(0,0);
    }
};
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