<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>

#include <vector>
#include <string>
#include <climits>
#include <algorithm>

using namespace std;

class Solution {
public:
    const int MOD = 1e9 + 7; // Use modulo for large values
    
    int getLengthOfOptimalCompression(string s, int k) {
        int n = s.size();
        // dp[i][k] means the minimum length for substring s[0:i] with at most k deletions
        vector<vector<long long>> dp(n + 1, vector<long long>(k + 1, LLONG_MAX));
        dp[0][0] = 0;  // Base case: no characters, no deletions, length is 0
        
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= k; ++j) {
                if (dp[i-1][j] == LLONG_MAX) continue;  // Skip unreachable states
                
                // Case 1: Delete s[i-1], i.e., delete the current character
                if (j < k) {
                    dp[i][j+1] = min(dp[i][j+1], dp[i-1][j]);
                }
                
                // Case 2: Keep s[i-1] and count the run of characters
                int same = 0, diff = 0;
                for (int l = i; l <= n; ++l) {
                    if (s[l-1] == s[i-1]) same++;  // Count the same characters
                    else diff++;  // Count different characters
                    
                    if (j + diff <= k) {
                        dp[l][j + diff] = min(dp[l][j + diff], dp[i-1][j] + 1 + (same >= 100 ? 3 : same >= 10 ? 2 : same >= 2 ? 1 : 0));
                    } else {
                        break;
                    }
                }
            }
        }
        
        // Find the minimum value in the last row
        long long ans = LLONG_MAX;
        for (int j = 0; j <= k; ++j) {
            ans = min(ans, dp[n][j]);
        }
        
        return static_cast<int>(ans % MOD);  // Ensure the result is within the required range
    }
};

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