<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:
#     def getLengthOfOptimalCompression(self, s: str, k: int) -> int:
#         chars = defaultdict(int)
#         ans = ""
#         for i in s:
#             chars[i] += 1
#         for key, val in chars.items():
#             if k >= val:
#                 k -= val
#                 chars[key] = 0
#             else:
#                 if val == 1:
#                     ans += key
#                 else:
#                     ans += key + str(val)
#         if k > 0:
#             maxVal = max(chars.values())
#             for key, val in chars.items():
#                 if val == maxVal:
#                     idx = ans.find(str(val))
#                     ans = ans[:idx] + str(val-1) + ans[idx+len(str(val)):]
#         print(ans)
#         return len(ans)
class Solution:
    def __init__(self):
        self.dic = {}

    def getLengthOfOptimalCompression(self , s : str , k : int) -> int:
        def recursive_function(curIdx , st , k , prev , freq):

            if(curIdx == len(s)):
                return 0
            
            key = (curIdx , k , prev , freq)

            if(key in self.dic):
                return self.dic[key]

            increment_count = (1) if((freq == 1) or (freq == 9) or (freq == 99)) else (0)

            pick = (increment_count + recursive_function((curIdx + 1) , st , k , prev , (freq + 1))) if(st[curIdx] == prev) else (1 + recursive_function((curIdx + 1) , st , k , st[curIdx] , 1))

            not_pick = (recursive_function((curIdx + 1) , st , (k - 1) , prev , freq)) if(k) else (float('inf'))


            self.dic[key] = min(pick , not_pick)

            return self.dic[key]
    
        return recursive_function(0 , s , k , '' , 0)       
360700