<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>> dp(n, vector<int>(n));
        int i=0,j=n-1;
        bool found;
        while (i<=j) {
            found = false;
            for (int p=i; p<(i+(j-i+1)/2); ++p) {
                if (text[p]!=text[j]) continue;
                if (text.substr(i, p-i+1) != text.substr(j-p+i, p-i+1)) continue;
                found = true;
                dp[p][j-p+i] = (i==0) ? 2 : dp[i-1][j+1] + 2;
                j = j-p+i-1; i = p+1; 
                break;
            }
            if (found) continue;
            dp[i][j] = (i==0) ? 1 : dp[i-1][j+1] + 1;
            return dp[i][j];
        }
        return dp[i-1][j+1];
    }
};
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