<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) {
      const int n = s.length();
    vector<vector<int>> cache(n, vector<int>(k + 1, -1));
    function<int(int, int)> dp = [&](int i, int k) -> int {
      if (k < 0) return n;
      if (i + k >= n) {
        if(i<n) return cache[i][k]=0;
        else return 0;
      }
      int& ans = cache[i][k];
      if (ans != -1) return ans;
      ans = dp(i + 1, k - 1); // delete      
      int len = 0;
      int same = 0;
      int diff = 0;
      for (int j = i; j < n && diff <= k; ++j) {
        if (s[j] == s[i] && ++same) {
          if (same <= 2 || same == 10 || same == 100) ++len;
        } else {
          ++diff;
        }
        ans = min(ans, len + dp(j + 1, k - diff));
      }
      return ans;
    };
    return dp(0, k);
        // int n = s.size();
       
        // vector<vector<int>> dp(n + 1, vector<int>(k + 1, INT_MAX));
        // for(int i=0; i<=k; i++)
        //     dp[0][i]=-i; 

        // auto lengthOfCount = [](int count) {
        //     if (count == 1) return 0;      
        //     else if (count < 10) return 1;  
        //     else if (count < 100) return 2; 
        //     else return 3;                  
        // };

        // for (int i = 1; i <= n; ++i) {
        //     for (int j = 0; j <= k; ++j) {
        //         if (dp[i-1][j] == INT_MAX) continue;

        //         int count = 0, deletions = 0;
        //         for (int x = i; x <= n; ++x) {
        //             if (s[x - 1] == s[i - 1]) {
        //                 count++;
        //             } else {
        //                 deletions++;
        //             }

        //             if (j + deletions <= k) {
        //                 dp[x][j + deletions] = min(dp[x][j + deletions], 
        //                                            dp[i - 1][j] + 1 + lengthOfCount(count));
        //             }
        //         }

        //         if (j + 1 <= k) {
        //             dp[i][j + 1] = min(dp[i][j + 1], dp[i-1][j]);
        //         }
        //     }
        // }

        // return dp[n][k];
    }
};
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