<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<tuple>
#include<unordered_map>
#include<unordered_set>


class Solution {
    using Index = int;
    using LastChar = int;
    using LastCharCount = int;
    using RemovalsLeft = int;
    using Key = long long;
    using CompressedLen = int;
    std::unordered_map<Key, CompressedLen> memo;
    std::unordered_set<LastCharCount> counts = {1, 9, 99};

    CompressedLen solve(const string& s, int index, char lastChar, int lastCharCount, int removalsLeft)
    {
        if(removalsLeft < 0) {
            return 1001;
        }

        const int size = s.size();
        if(index == size) return 0;

        Key nodeKey = removalsLeft + (lastCharCount << 8) + (lastChar <<  16) + (index << 24);
        if(memo.find(nodeKey) != memo.cend()) {
            return memo.at(nodeKey);
        }

        int removedCount = solve(s, index + 1, lastChar, lastCharCount, removalsLeft - 1);
        int keptCount = 0;

        if(s[index] == lastChar) {
            const bool doesDigitChange = counts.find(lastCharCount) != counts.cend();
            keptCount = solve(s, index + 1, lastChar, lastCharCount + 1, removalsLeft) + (doesDigitChange ? 1 : 0); 
        }
        else
        {
            keptCount = solve(s, index + 1, s[index], 1, removalsLeft) + 1;
        }

        const CompressedLen ret = std::min(keptCount, removedCount);
        memo[nodeKey] = ret;

        return ret;
    }

public:

    int getLengthOfOptimalCompression(const string& s, int k) {
        return solve(s, 0, 0, 0, k);
    }
};
375400