<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 getLengthOfOptimalCompression(string s, int k) {
        int n = s.length();
        vector<vector<int>> dp(110, vector<int>(110, 9999)); // Initializing a 2D vector 'dp' of size 110x110 with value 9999
        dp[0][0] = 0; // Initializing the base case where no characters and deletions exist

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                int cnt = 0, del = 0;
                for (int l = i; l >= 1; l--) {
                    // Count the frequency of characters from 'i' to 'l'
                    if (s[l - 1] == s[i - 1]) 
                        cnt++;
                    else 
                        del++;
                    
                    // Check if the remaining allowed deletions are valid (j - del >= 0)
                    if (j - del >= 0) {
                        // Update the dp array based on the conditions
                        dp[i][j] = min(dp[i][j],
                            dp[l - 1][j - del] + 1 + (cnt == 100 ? 3 : cnt >= 10 ? 2 : cnt >= 2 ? 1 : 0));
                    }
                }
                
                // If there are remaining allowed deletions (j > 0), consider the case without deleting current character
                if (j > 0)
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
            }
        }
        return dp[n][k]; // Return the minimum length for 's' with at most 'k' deletions
    }
};



17400