<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 dp[1001][1001];
    int solve(int i, int j, string& text){
        if(i>j) return 0;
        else if(i == j) return 1;
        else if(dp[i][j] != -1) return dp[i][j];
        else{
            int res = 1;
            // cout<<endl;
            // cout<<i<<" "<<j<<endl;
            // cout<<endl;
            int l = 0, r = 0;
            while(i+l<=j-r){
                // cout<<l<<" "<<r<<endl;
                // cout<<text.substr(i,l+1)<<" "<<text.substr(j-r,r+1)<<endl;
                if(text.substr(i,l+1) == text.substr(j-r,r+1)) {
                    // cout<<"here"<<endl;
                    res = max(res,2+solve(i+l+1,j-r-1,text));
                }
                l++;
                r++;
            }
            return dp[i][j] = res;
        }
    }
    int longestDecomposition(string text) {
        memset(dp,-1,sizeof(dp));
        int n = text.size();
        return solve(0,n-1,text);
    }
};
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