<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:
bool word(vector<string> &v, int nv, string s, int ns, string temp)
{
    if (temp.size() > ns)
    {
        return false;
    }
    if (temp == s)
    {
        return true;
    }
    bool ans = false;
    for (int i = 0; i < nv; i++)
    {
        ans = (ans || word(v, nv, s, ns, temp + v[i]));
    }
    return ans;
}
bool dpword(vector<string> &v, int nv, string s, int ns)
{
    vector<bool> dp(ns + 1);
    dp[0] = true;
    unordered_map<string, bool> mp;
    for (int i = 0; i < nv; i++)
    {
        mp[v[i]] = true;
    }
    for (int i = 1; i <= ns; i++)
    {
        for (int j = 0; j < i; j++)
        {
            string subs = "";
            for (int k = j; k < i; k++)
            {
                subs += s[k];
            }

            if (dp[j] && (mp.find(subs) != mp.end()))
            {
                dp[i] = true;
                break;
            }
        }
    }

    return dp[ns];
}
    bool wordBreak(string s, vector<string>& v) {
        int nv = v.size();
        int ns = s.size();
        // string temp = "";
        // return word(v,nv,s,ns,temp);
        return dpword(v,nv,s,ns);
    }
};
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