<p>You are given a string <code>text</code>. You should split it to k substrings <code>(subtext<sub>1</sub>, subtext<sub>2</sub>, ..., subtext<sub>k</sub>)</code> such that:</p>

<ul>
	<li><code>subtext<sub>i</sub></code> is a <strong>non-empty</strong> string.</li>
	<li>The concatenation of all the substrings is equal to <code>text</code> (i.e., <code>subtext<sub>1</sub> + subtext<sub>2</sub> + ... + subtext<sub>k</sub> == text</code>).</li>
	<li><code>subtext<sub>i</sub> == subtext<sub>k - i + 1</sub></code> for all valid values of <code>i</code> (i.e., <code>1 &lt;= i &lt;= k</code>).</li>
</ul>

<p>Return the largest possible value of <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;ghiabcdefhelloadamhelloabcdefghi&quot;
<strong>Output:</strong> 7
<strong>Explanation:</strong> We can split the string on &quot;(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;merchant&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We can split the string on &quot;(merchant)&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;antaprezatepzapreanta&quot;
<strong>Output:</strong> 11
<strong>Explanation:</strong> We can split the string on &quot;(a)(nt)(a)(pre)(za)(tep)(za)(pre)(a)(nt)(a)&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text.length &lt;= 1000</code></li>
	<li><code>text</code> consists only of lowercase English characters.</li>
</ul>

class Solution {
public:
    int helper(string s, unordered_map<string, int> &map)
    {
        if(s.size()==0)
        {
            return 0;
        }
        if(map.find(s) != map.end())
        {
            return map[s];
        }
        int ans=1;
        for(int i=0;i<s.size()/2;i++)
        {
            if( s.substr(0,i+1)==s.substr(s.size()-i-1) )
            {
                // int x = helper( s.substr(i+1, s.size()-2*i ) , map);
                // cout<<s.substr(0,i+1)<<" "<< s.substr(s.size()-i-1)<<endl;
                ans=max( ans , 2 + helper( s.substr(i+1, s.size()-2*(i+1) ) , map) );
            }
        }
        map[s] = ans;
        return ans;
    }
    int longestDecomposition(string s) {
        unordered_map<string, int> map;
        return helper(s, map);
    }
};
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