<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 {
    vector<vector<int>> dp;
    
    int Contribution (int cnt) {
        if (cnt == 1) return 0;
        return to_string(cnt).length();
    } 
    
    int LengthOfOptimalCompression (int ind, int k, const string& s) {
        int n = s.length();
        if (ind >= n) return 0;
        
        int& ans = dp[ind][k];
        if (ans != -1) return ans;
        ans = 1e9;
        
        int equal = 0, removed = 0;
        for (int nxt_valid_ind = ind; removed <= k && nxt_valid_ind < n; nxt_valid_ind ++) {
            if (s[nxt_valid_ind] == s[ind]) {
                equal ++;
                continue;
            }
            
            for (int eq_rmv = 0; eq_rmv <= min (equal, k-removed); eq_rmv ++) {
                int cur_lft = equal - eq_rmv;
                int rmv_lft = k - eq_rmv - removed;
                ans = min (
                    ans, 
                    1 + Contribution(cur_lft) + LengthOfOptimalCompression(nxt_valid_ind, rmv_lft, s)
                );
            }
            removed ++;
        }
        
        if (removed <= k) {
            ans = min (ans, 1 + Contribution(equal - (k - removed)));
        }
        
        return ans;
    }
    
public:
    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.length();
        dp.resize(n, vector<int>(k+1, -1));
        
        int result = 1e9;
        for (int j = 0; j <= k; j ++) {
            result = min (result, LengthOfOptimalCompression(j, k-j, s));
        }
        return result;
    }
};
13400