<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 recursive(string& s, int n, int start, int k, vector<vector<int>>& dp) {
        // Base case: if k is negative, the subproblem is invalid.
        if (k < 0) return n;
        
        // Base case: if we've reached the end of the string, no more compression needed.
        if (start == n) return 0;
        
        // Base case: if only one character remains.
        if (start == n - 1) return k == 0 ? 1 : 0;
        
        // Base case: if the remaining part of the string can be deleted.
        if (n - start <= k) return 0;
        
        // Use memoization to avoid recomputing subproblems.
        if (dp[start][k] != -1) return dp[start][k];
        
        // Array to count the occurrences of each character in the substring.
        vector<int> cnt(26, 0);
        
        // Variable to store the maximum occurrence of any character in the current substring.
        int maxCnt = 0;
        
        // Initialize the answer to a large value.
        int ans = n;
        
        // Iterate through the string from the current start position.
        for (int i = start; i < n; ++i) {
            // Increment the count of the current character.
            maxCnt = max(maxCnt, ++cnt[s[i] - 'a']);
            
            // Calculate the length of the current substring.
            int strLen = i - start + 1;
            
            // Calculate the number of characters to delete to get maxCnt identical characters.
            int cntToDelete = strLen - maxCnt;
            
            // Update the answer using the formula from the pseudocode.
            int digits_of_maxCnt = (maxCnt > 1) ? static_cast<int>(to_string(maxCnt).length()) : 0;
            ans = min(ans, 1 + digits_of_maxCnt + recursive(s, n, i + 1, k - cntToDelete, dp));
        }
        
        // Store the result in dp array and return.
        return dp[start][k] = ans;
    }
    
    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.length();
        vector<vector<int>> dp(n, vector<int>(k + 1, -1));
        return recursive(s, n, 0, k, dp);
    }
};

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