<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:
    
    int getLength(int count){

        // suppose a is coming 9 times , so we write a9 - length(a9) = 2
        // count>=10 - suppose 15 times , so length(a15) string = 3
        
        if(count==1)return 1; 
        else{
            if(count<10)return 2; 
            else{
                if(count<100)return 3; 
                else return 4;  
            }
        }
    }
    
    
    int getLengthOfOptimalCompression(string s, int k) {
        
        int n = s.size();
        
        int dp[n+1][k+1];
        
        for(int i=n;i>=0;i--){
            for(int j=0;j<=k;j++){
                
                // declare the base case - initial length declare
                if(i == n) { dp[n][j] = 0;  continue;}
                
                // case 1 - delete the ith character if possible (j>0)
                dp[i][j] = j>0 ? dp[i+1][j-1]:INT_MAX;
                
                // case 2-  we keep the ith index.
                
                //we have to find the current char's index end point
                
                int possible_del = j,count=0;
                for(int end =i;end<n&&possible_del>=0 ;end++){
                    // check about the same char "b b b .." for encoding
                    if(s[end] == s[i]){
                        count++;
                        
                        // assumeing the block end here "aaaa"
                        dp[i][j] = min(dp[i][j] , getLength(count)+dp[end+1][possible_del]);
                    }
                    else { 
                        // this character should be deleted. to make length min.
                        possible_del--;
                    }
                }
                
                
            }
        }
        
        return dp[0][k];
    }
};
7800