<p>You are given a string <code>s</code> consisting of lowercase letters and an integer <code>k</code>. We call a string <code>t</code> <strong>ideal</strong> if the following conditions are satisfied:</p>

<ul>
	<li><code>t</code> is a <strong>subsequence</strong> of the string <code>s</code>.</li>
	<li>The absolute difference in the alphabet order of every two <strong>adjacent</strong> letters in <code>t</code> is less than or equal to <code>k</code>.</li>
</ul>

<p>Return <em>the length of the <strong>longest</strong> ideal string</em>.</p>

<p>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</p>

<p><strong>Note</strong> that the alphabet order is not cyclic. For example, the absolute difference in the alphabet order of <code>&#39;a&#39;</code> and <code>&#39;z&#39;</code> is <code>25</code>, not <code>1</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;acfgbd&quot;, k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest ideal string is &quot;acbd&quot;. The length of this string is 4, so 4 is returned.
Note that &quot;acfgbd&quot; is not ideal because &#39;c&#39; and &#39;f&#39; have a difference of 3 in alphabet order.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, k = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest ideal string is &quot;abcd&quot;. The length of this string is 4, so 4 is returned.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= k &lt;= 25</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def longestIdealString(self, s: str, k: int) -> int:
        n = len(s)

        # Initialize all dp values to -1 to indicate non-visited states
        dp = [[-1] * 26 for _ in range(n)]

        def dfs(i: int, c: int) -> int:
            # Memoized value
            if dp[i][c] != -1:
                return dp[i][c]

            # State is not visited yet
            match = c == (ord(s[i]) - ord('a'))
            dp[i][c] = 1 if match else 0

            # Non base case handling
            if i > 0:
                dp[i][c] = dfs(i - 1, c)
                if match:
                    for p in range(26):
                        if abs(c - p) <= k:
                            dp[i][c] = max(dp[i][c], dfs(i - 1, p) + 1)
            return dp[i][c]

        # Find the maximum dp[N-1][c] and return the result
        res = 0
        for y in range(26):
            res = max(res, dfs(n - 1, y))
        return res
72300