<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 getLengthOfOptimalCompression(string s, int k) {
    vector<vector<int>> mem(s.length(), vector<int>(k + 1, kMax));
    return compression(s, 0, k, mem);
  }

 private:
  static constexpr int kMax = 101;

  // Returns the length of the optimal compression of s[i..n) with at most k
  // deletion.
  int compression(const string& s, int i, int k, vector<vector<int>>& mem) {
    if (k < 0)
      return kMax;
    if (i == s.length() || s.length() - i <= k)
      return 0;
    if (mem[i][k] != kMax)
      return mem[i][k];

    int maxFreq = 0;  // the maximum frequency in s[i..j]
    vector<int> count(128);

    // Make letters in s[i..j] be the same.
    // Keep the letter that has the maximum frequency in this range and remove
    // the other letters.
    for (int j = i; j < s.length(); ++j) {
      maxFreq = max(maxFreq, ++count[s[j]]);
      mem[i][k] = min(  //
          mem[i][k],    //
          getLength(maxFreq) +
              compression(s, j + 1, k - (j - i + 1 - maxFreq), mem));
    }

    return mem[i][k];
  }

  // Returns the length to compress `maxFreq`.
  int getLength(int maxFreq) {
    if (maxFreq == 1)
      return 1;  // c
    if (maxFreq < 10)
      return 2;  // [1-9]c
    if (maxFreq < 100)
      return 3;  // [1-9][0-9]c
    return 4;    // [1-9][0-9][0-9]c
  }
};
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