Longest Chunked Palindrome Decomposition
<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>
Greedy; Two Pointers; String; Dynamic Programming; Hash Function; Rolling Hash
function longestDecomposition(text: string): number {
    let ans = 0;
    for (let i = 0, j = text.length - 1; i <= j; ) {
        let ok = false;
        for (let k = 1; i + k - 1 < j - k + 1; ++k) {
            if (text.slice(i, i + k) === text.slice(j - k + 1, j + 1)) {
                ans += 2;
                i += k;
                j -= k;
                ok = true;
                break;
            }
        }
        if (!ok) {
            ++ans;
            break;
        }
    }
    return ans;
}
