<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 {
public:
    int longestIdealString(string s, int k) {
        unordered_map<char, int> ump;
        for (char c = 'a'; c <= 'z'; ++c)
            ump[c] = c - 'a';
        
        int n = s.length();
        vector<vector<int>> dp(n + 1, vector<int>(27, 0));

        for (int ind = n - 1; ind >= 0; --ind) {
            for (int prev = 0; prev <= 26; ++prev) {
                int val1 = dp[ind + 1][prev];
                int val2 = 0;
                if (prev == 26 || abs(ump[prev + 'a'] - ump[s[ind]]) <= k)
                    val2 = 1 + dp[ind + 1][s[ind] - 'a'];
                dp[ind][prev] = max(val1, val2);
            }
        }

        return dp[0][26]; // Initial call with prevchar as '\0' equivalent
    }
};
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