Maximum Deletions on a String
<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>
String; Dynamic Programming; String Matching; Hash Function; Rolling Hash
class Solution {
public:
    int deleteString(string s) {
        int n = s.size();
        int g[n + 1][n + 1];
        memset(g, 0, sizeof(g));
        for (int i = n - 1; ~i; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    g[i][j] = g[i + 1][j + 1] + 1;
                }
            }
        }
        int f[n];
        memset(f, 0, sizeof(f));
        function<int(int)> dfs = [&](int i) -> int {
            if (i == n) {
                return 0;
            }
            if (f[i]) {
                return f[i];
            }
            f[i] = 1;
            for (int j = 1; j <= (n - i) / 2; ++j) {
                if (g[i][i + j] >= j) {
                    f[i] = max(f[i], 1 + dfs(i + j));
                }
            }
            return f[i];
        };
        return dfs(0);
    }
};