<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>

import math
class Solution:
    def deleteString(self, s: str) -> int:
        self.dp = [-1]*len(s)

        return self.delStrUtil(s,0)

    def delStrUtil(self,s,i):
        if self.dp[i]!=-1:
            return self.dp[i]
        n = len(s[i:])
        maxFilterSize = (n//2)*2
        size = 2

        # print(s,'1')
        maxCount = 1

        while size<=maxFilterSize:
            if self.isEqual(s[i:i+size],size):
                # print(s[i:i+size])
                maxCount = max(maxCount,1 + self.delStrUtil(s,i+size//2))
            size+=2
        
        self.dp[i] = maxCount
        # print(s,'3')
        return maxCount

    def isEqual(self,s,n):
        return s[:n//2]==s[n//2:]

7055