<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)
        K = k

        # Helper function to compute the length added by a run of a certain count
        def get_length(count):
            if count == 1:
                return 1  # Only the character
            elif count < 10:
                return 2  # Character + one-digit count
            elif count < 100:
                return 3  # Character + two-digit count
            else:
                return 4  # Character + three-digit count

        # Initialize DP array
        dp = [[float('inf')] * (K + 1) for _ in range(N + 1)]
        for k in range(K + 1):
            dp[N][k] = 0  # Base case: empty string

        # Fill DP table
        for i in range(N - 1, -1, -1):
            for k in range(K + 1):
                # Option 1: Delete s[i]
                if k > 0:
                    dp[i][k] = min(dp[i][k], dp[i + 1][k - 1])

                # Option 2: Keep s[i] and try to build runs
                cnt = 0  # Count of s[i]
                num_deletions = 0
                char_count = [0] * 26  # Frequency of each character

                for j in range(i, N):
                    idx = ord(s[j]) - ord('a')
                    char_count[idx] += 1

                    if s[j] == s[i]:
                        cnt += 1
                    else:
                        num_deletions += 1

                    if num_deletions > k:
                        break

                    # Calculate compressed length for the current run
                    compressed_len = get_length(cnt)
                    dp[i][k] = min(dp[i][k], compressed_len + dp[j + 1][k - num_deletions])

        return dp[0][K]

17000