<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 {
private:
    static constexpr int MAX = 127;

    int numEncodedDigits(int count) {
        if (count <= 1) {
            return 0;
        } else if (count < 10) {
            return 1;
        } else if (count < 100) {
            return 2;
        }
        return 3;
    }

    int minRLEString(const string& s, int start, int k, vector<vector<int>>& dp) {
        if (k < 0) {
            return MAX;
        }
        
        int n = s.size();
        if (start >= n || n - start <= k) {
            return 0;
        }

        if (dp[start][k] != -1) {
            return dp[start][k];
        }

        dp[start][k] = n;
        array<int, 26> letter_freq{};
        int max_freq = 0;

        for (int i = start; i < n; i++) {
            max_freq = max(max_freq, ++letter_freq[s[i] - 'a']);
            int num_delete = i - start + 1 - max_freq;
            dp[start][k] = min(dp[start][k], 1 + numEncodedDigits(max_freq) + minRLEString(s, i + 1, k - num_delete, dp));
        }

        return dp[start][k];
    }

public:
    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(k + 1, -1));
        
        return minRLEString(s, 0, k, dp);
    }
};
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