<p>You are given a string <code>s</code> consisting of only lowercase English letters. In one operation, you can:</p>

<ul>
	<li>Delete <strong>the entire string</strong> <code>s</code>, or</li>
	<li>Delete the <strong>first</strong> <code>i</code> letters of <code>s</code> if the first <code>i</code> letters of <code>s</code> are <strong>equal</strong> to the following <code>i</code> letters in <code>s</code>, for any <code>i</code> in the range <code>1 &lt;= i &lt;= s.length / 2</code>.</li>
</ul>

<p>For example, if <code>s = &quot;ababc&quot;</code>, then in one operation, you could delete the first two letters of <code>s</code> to get <code>&quot;abc&quot;</code>, since the first two letters of <code>s</code> and the following two letters of <code>s</code> are both equal to <code>&quot;ab&quot;</code>.</p>

<p>Return <em>the <strong>maximum</strong> number of operations needed to delete all of </em><code>s</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abcabcdabc&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong>
- Delete the first 3 letters (&quot;abc&quot;) since the next 3 letters are equal. Now, s = &quot;abcdabc&quot;.
- Delete all the letters.
We used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed.
Note that in the second operation we cannot delete &quot;abc&quot; again because the next occurrence of &quot;abc&quot; does not happen in the next 3 letters.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaabaab&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong>
- Delete the first letter (&quot;a&quot;) since the next letter is equal. Now, s = &quot;aabaab&quot;.
- Delete the first 3 letters (&quot;aab&quot;) since the next 3 letters are equal. Now, s = &quot;aab&quot;.
- Delete the first letter (&quot;a&quot;) since the next letter is equal. Now, s = &quot;ab&quot;.
- Delete all the letters.
We used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaaaa&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> In each operation, we can delete the first letter of s.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 4000</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

class StringHash:
    A = 911382323
    B = 972663749

    def __init__(self, s):
        self.h = [0] * len(s)
        self.p = [0] * len(s)
        self.h[0] = ord(s[0])
        self.p[0] = 1
        for i in range(1, len(s)):
            self.h[i] = (self.h[i - 1] * self.A + ord(s[i])) % self.B
            self.p[i] = (self.p[i - 1] * self.A) % self.B

    def calc(self, l, r):
        if l == 0:
            return self.h[r]
        return (self.h[r] - self.h[l - 1] * self.p[r - l + 1]) % self.B

class Solution:
    def deleteString(self, s: str) -> int:
        h = StringHash(s)
        @cache
        def dp(i):
            if i >= len(s): return False
            res = 1
            for j in range(i, len(s)):
                l = j - i + 1
                if 2 * l > (len(s) - i): break
                if h.calc(i, j) == h.calc(j + 1, j + l): res = max(res, 1 + dp(j + 1))
            return res
        return dp(0)
22500