<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 longestDecomposition(string text)
    {
        int n=text.length();
        vector<vector<int>>maxc(n,vector<int>(n));
        for(int i=0;i<n;i++)maxc[i][i]=1;
        for(int i=n-2;i>=0;i--)for(int j=i+1;j<n;j++)
        {
            if(text[i]!=text[j])maxc[i][j]=0;
            else if(j!=n-1)maxc[i][j]=1+maxc[i+1][j+1];
            else maxc[i][j]=1;
        }
        int l=0,r=n-1,e=n-1,ans=0;
        while(l<=r)
        {
            if(l==r)return ++ans;
            if(maxc[l][r]<e-r+1)r--;
            else{
                int len=e-r+1;
                if(2*len>(e-l+1))return ++ans;
                if(2*len==(e-l+1)){ans+=2;return ans;}
                ans+=2;l+=len;e-=len;r=e;
            }
        }return ans;
    }
};
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