<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:
    int helper(int index,string s,vector<string>& words,vector<int>&dp)
    {
        if(index==s.length())
        return 0;

        if(dp[index]!=-1)
        return dp[index];

        int ans=301;
        int n=words.size();
        int sz=s.length();

        for(int i=0;i<n;i++)
        {
            bool present=true;  
            if(sz-index<words[i].length())
            continue;

            for(int j=0;j<words[i].length();j++)
            {
                if(s[index+j]!=words[i][j])
                {
                    present=false;
                    break;
                }
            }
            if(present)
            {
                ans=min(ans,1+helper(index+words[i].length(),s,words,dp));
            }
            
        }
        return dp[index]=ans;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        int sz=s.length();
        int n=wordDict.size();

        vector<int> dp(sz,-1);

        int val=helper(0,s,wordDict,dp);
        return val>=301?false:true;
    }
};
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