<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>

int dp[101][27][101][101]; 

class Solution {
public:

  
    int solveMem(string& s, int index, int prev, int k, int count){
        // Base case: if we have used more deletions than allowed
        if(k < 0) return 1e9;

        // Base case: if we have reached the end of the string
        if(index >= s.size()){
            return 0;
        }

        // If the result for the current state is already computed, return it
        if(dp[index][prev][k][count] != -1){
            return dp[index][prev][k][count];
        }

        // Initialize the accumulated cost to a large value
        int acc = INT_MAX;

        // Recursive call for deleting the current character
        int del = solveMem(s, index + 1, prev, k - 1, count);

        // Check if the current character is the same as the previous one
        // Check if the current character is the same as the previous one
        if(prev + 'a' == s[index]){
            // If the count of consecutive characters is 1, 9, or 99
            if(count == 1 || count == 9 || count == 99){
                // Update the accumulated cost by 1 and recursively call the function
                acc = 1 + solveMem(s, index + 1, prev, k, count + 1);
            }
            else{
                // If the count is not 1, 9, or 99, recursively call the function without updating the cost
                acc = solveMem(s, index + 1, prev, k, count + 1);
            }
        }

        else{
            // If the current character is different, update the accumulated cost accordingly
            acc = 1 + solveMem(s, index + 1, s[index] - 'a', k, 1);
        }

        // Store the minimum of the delete and accumulate operations in the DP table
        return dp[index][prev][k][count] = min(del, acc);
    }


    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.size();

        // Initialize the DP table with -1
        memset(dp, -1, sizeof(dp));

        // Call the recursive function with initial parameters
        return solveMem(s, 0, 26, k, 0);
    }
};
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