Maximize the Number of Partitions After Operations
<p>You are given a string <code>s</code> and an integer <code>k</code>.</p>

<p>First, you are allowed to change <strong>at most</strong> <strong>one</strong> index in <code>s</code> to another lowercase English letter.</p>

<p>After that, do the following partitioning operation until <code>s</code> is <strong>empty</strong>:</p>

<ul>
	<li>Choose the <strong>longest</strong> <strong>prefix</strong> of <code>s</code> containing at most <code>k</code> <strong>distinct</strong> characters.</li>
	<li><strong>Delete</strong> the prefix from <code>s</code> and increase the number of partitions by one. The remaining characters (if any) in <code>s</code> maintain their initial order.</li>
</ul>

<p>Return an integer denoting the <strong>maximum</strong> number of resulting partitions after the operations by optimally choosing at most one index to change.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;accca&quot;, k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>The optimal way is to change <code>s[2]</code> to something other than a and c, for example, b. then it becomes <code>&quot;acbca&quot;</code>.</p>

<p>Then we perform the operations:</p>

<ol>
	<li>The longest prefix containing at most 2 distinct characters is <code>&quot;ac&quot;</code>, we remove it and <code>s</code> becomes <code>&quot;bca&quot;</code>.</li>
	<li>Now The longest prefix containing at most 2 distinct characters is <code>&quot;bc&quot;</code>, so we remove it and <code>s</code> becomes <code>&quot;a&quot;</code>.</li>
	<li>Finally, we remove <code>&quot;a&quot;</code> and <code>s</code> becomes empty, so the procedure ends.</li>
</ol>

<p>Doing the operations, the string is divided into 3 partitions, so the answer is 3.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;aabaab&quot;, k = 3</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p>Initially&nbsp;<code>s</code>&nbsp;contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;xxyz&quot;, k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The optimal way is to change&nbsp;<code>s[0]</code>&nbsp;or&nbsp;<code>s[1]</code>&nbsp;to something other than characters in&nbsp;<code>s</code>, for example, to change&nbsp;<code>s[0]</code>&nbsp;to&nbsp;<code>w</code>.</p>

<p>Then&nbsp;<code>s</code>&nbsp;becomes <code>&quot;wxyz&quot;</code>, which consists of 4 distinct characters, so as <code>k</code> is 1, it will divide into 4 partitions.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= 26</code></li>
</ul>
Bit Manipulation; String; Dynamic Programming; Bitmask
class Solution:
    def maxPartitionsAfterOperations(self, s: str, k: int) -> int:
        @cache
        def dfs(i: int, cur: int, t: int) -> int:
            if i >= n:
                return 1
            v = 1 << (ord(s[i]) - ord("a"))
            nxt = cur | v
            if nxt.bit_count() > k:
                ans = dfs(i + 1, v, t) + 1
            else:
                ans = dfs(i + 1, nxt, t)
            if t:
                for j in range(26):
                    nxt = cur | (1 << j)
                    if nxt.bit_count() > k:
                        ans = max(ans, dfs(i + 1, 1 << j, 0) + 1)
                    else:
                        ans = max(ans, dfs(i + 1, nxt, 0))
            return ans

        n = len(s)
        return dfs(0, 0, 1)
