<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 Solution:
    def deleteString(self, s: str) -> int:
        from collections import Counter
        N = len(s)
        global memo
        memo = {}
        #print('\n\nTest case -> s:', s, 'N:', N)
        return rem(s)


def rem(s):
    #print('Rem Start -> s:', s)
    global memo
    if s in memo:
        #print('MEMO found -> memo[s]:', memo[s])
        return memo[s]
    
    N = len(s)
    #print('N:', N)
    if N == 0:
        #print('Empty string case, s:', s)
        return 0
    
    if N == 1:
        # Remove entire string
        #print('Single char case, s:', s)
        memo[s] = 1
        return 1
    
    c = Counter(s)
    if c[s[0]] == N:
        # All chars are same
        memo[s] = N
        return N
    
    maxToRem = N // 2
    #print('maxToRem:', maxToRem)
    maxSteps = 1
    for numToRem in range(maxToRem, 0, -1):
        if s[:numToRem] == s[numToRem:2*numToRem]:
            #numCharsToRem,append(numToRem)
            #print('s:', s, 'numToRem:', numToRem, 'Removing', s[:numToRem])
            maxSteps = max(maxSteps, rem(s[numToRem:])+1)
    
    memo[s] = maxSteps
    #print('s:', s, 'memo[s]:', memo[s])
    return maxSteps
23000