<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) {

       long long n=s.size();
       vector<long long> dp(n+1,0);
    
       map<long long,long long> mp;

       long long ans=-1;
       for(int i=1;i<=n;i++){
             for(int j=(-k);j<=k;j++){
                if(mp[abs((s[i-1]-'a')-j)]>0){
                    // cout<<"in="<<mp[abs((s[i-1]-'a')-k)]<<endl;
                    // cout<<i<<endl;
                   dp[i]=max(dp[i],(dp[mp[abs((s[i-1]-'a')-j)]]+1));
                }
             }
             mp[(s[i-1]-'a')]=i;
             ans=max(ans,dp[i]);
       }
    //    for(auto x:mp){
    //     cout<<x.first<<" "<<x.second<<endl;
    //    }
    //      for(auto x:dp){
    //         cout<<x<<" ";
    //      }
    //      cout<<endl;
       return (ans+1);







    }
};
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