<p><a href="http://en.wikipedia.org/wiki/Run-length_encoding">Run-length encoding</a> is a string compression method that works by&nbsp;replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string&nbsp;<code>&quot;aabccc&quot;</code>&nbsp;we replace <font face="monospace"><code>&quot;aa&quot;</code></font>&nbsp;by&nbsp;<font face="monospace"><code>&quot;a2&quot;</code></font>&nbsp;and replace <font face="monospace"><code>&quot;ccc&quot;</code></font>&nbsp;by&nbsp;<font face="monospace"><code>&quot;c3&quot;</code></font>. Thus the compressed string becomes <font face="monospace"><code>&quot;a2bc3&quot;</code>.</font></p>

<p>Notice that in this problem, we are not adding&nbsp;<code>&#39;1&#39;</code>&nbsp;after single characters.</p>

<p>Given a&nbsp;string <code>s</code>&nbsp;and an integer <code>k</code>. You need to delete <strong>at most</strong>&nbsp;<code>k</code> characters from&nbsp;<code>s</code>&nbsp;such that the run-length encoded version of <code>s</code>&nbsp;has minimum length.</p>

<p>Find the <em>minimum length of the run-length encoded&nbsp;version of </em><code>s</code><em> after deleting at most </em><code>k</code><em> characters</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aaabcccd&quot;, k = 2
<strong>Output:</strong> 4
<b>Explanation: </b>Compressing s without deleting anything will give us &quot;a3bc3d&quot; of length 6. Deleting any of the characters &#39;a&#39; or &#39;c&#39; would at most decrease the length of the compressed string to 5, for instance delete 2 &#39;a&#39; then we will have s = &quot;abcccd&quot; which compressed is abc3d. Therefore, the optimal way is to delete &#39;b&#39; and &#39;d&#39;, then the compressed version of s will be &quot;a3c3&quot; of length 4.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aabbaa&quot;, k = 2
<strong>Output:</strong> 2
<b>Explanation: </b>If we delete both &#39;b&#39; characters, the resulting compressed string would be &quot;a4&quot; of length 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaaaaaaaaaa&quot;, k = 0
<strong>Output:</strong> 3
<strong>Explanation: </strong>Since k is zero, we cannot delete anything. The compressed string is &quot;a11&quot; of length 3.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>0 &lt;= k &lt;= s.length</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    unordered_map<int,int>mp;
    int f(string &s, int k,int i,int prev_char,int freq){
    
    if(k<0)return 1e9;
    if(i==s.length()){

       if(freq==0){
        return 0;
       }
       if(freq==1){
         return 1;
       } 
       if(freq>=2&&freq<=9){
        return 2;
       }
       if(freq>=10&&freq<=99){
        return 3;
       }
       return 4;
    }
    int mask=0;
    mask|=k;
    int j=9;
    int p=i;
    while(p>0){
     int m=p%2;
     p/=2;
     mask|=(m<<j);
     j++;
    }
    j=18;
    int p1=prev_char;
    while(p1>0){
     int m1=p1%2;
     p1/=2;
     mask|=(m1<<j);
     j++;
    }
    j=23;
    int p2=freq;
    while(p2>0){
     int m2=p2%2;
     p2/=2;
     mask|=(m2<<j);
     j++;
    }
    

    if(mp.find(mask)!=mp.end())return mp[mask];

    int ans=1e9;
    ans=min(ans,f(s,k-1,i+1,prev_char,freq));

    if(prev_char==-1||s[i]-'a'==prev_char){
      ans=min(ans,f(s,k,i+1,s[i]-'a',freq+1));
    }
    else{
      if(freq==1){
       ans=min(ans,1+f(s,k,i+1,s[i]-'a',1));
      }
      else if(freq>=2&&freq<=9){
       ans=min(ans,2+f(s,k,i+1,s[i]-'a',1));
      }
      else if(freq>=10&&freq<=99){
       ans=min(ans,3+f(s,k,i+1,s[i]-'a',1));
      }
      else if(freq==100){
        ans=min(ans,4+f(s,k,i+1,s[i]-'a',1));
      }
     
    }

    return mp[mask]=ans;



    }

    int getLengthOfOptimalCompression(string s, int k) {
        
        
        return f(s,k,0,-1,0);
    }
};
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