<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>

int dp[101][27][101][101];
class Solution {
public:
    int dcnt(int x){
        if(x==0)
            return -1;
        if(x==1)
            return 0;
        int res = 0;
        while(x){
            x/=10;
            res++;
        }
        return res;

    }
    int solve(int i,int k,int j,int c,vector<int> &v,vector<int> &u){
        if(i==v.size())
                return dcnt(j)+1;
        if(dp[i][c][j][k]!=-1)
            return dp[i][c][j][k];
        int ans = 1e9;
        for(int t = 0;t<=min(v[i],k);t++)
        {
            if(t==v[i])
                ans = min(ans,solve(i+1,k-t,j,c,v,u));
            else if(u[i]==c)
                ans = min(ans,solve(i+1,k-t,j+v[i]-t,c,v,u));
            else
                ans = min(ans,1+dcnt(j)+solve(i+1,k-t,v[i]-t,u[i],v,u));
        }
        return dp[i][c][j][k] = ans;

    }
    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.length();
        vector<int> v,u;
        for(int i = 0;i<n;i++){
            int j = i;
            while(j<n && s[j]==s[i])
                j++;
            v.push_back(j-i);
            u.push_back(s[i]-'a');
            i = j-1;
        }
        memset(dp,-1,sizeof(dp));
        return solve(0,k,0,26,v,u);
    }
};
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