<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:
        n = len(s)
        # Initialize a 2D list 'dp' with large values (9999) to represent the minimum length
        # for substrings with at most 'k' deletions.
        dp = [[9999] * 110 for _ in range(110)]
        dp[0][0] = 0  # Base case: no characters and no deletions result in length 0.

        # Iterate over each character in the string
        for i in range(1, n + 1):
            # Iterate over the number of deletions allowed
            for j in range(0, k + 1):
                cnt, del_ = 0, 0  # Initialize count of identical characters and deletions

                # Check the substring ending at the current character
                for l in range(i, 0, -1):
                    if s[l - 1] == s[i - 1]:
                        cnt += 1  # Increment count if characters match
                    else:
                        del_ += 1  # Increment deletions if characters do not match

                    # If the number of deletions is within the allowed limit
                    if j - del_ >= 0:
                        # Update the dp array with the minimum length
                        dp[i][j] = min(dp[i][j], dp[l - 1][j - del_] + 1 + (3 if cnt >= 100 else 2 if cnt >= 10 else 1 if cnt >= 2 else 0))

                # Consider the case where the current character is deleted
                if j > 0:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])

        # Return the minimum length for the entire string with at most 'k' deletions
        return dp[n][k]
16700