mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
71 lines
3.3 KiB
HTML
71 lines
3.3 KiB
HTML
<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> </p>
|
|
<p><strong class="example">Example 1:</strong></p>
|
|
|
|
<div class="example-block">
|
|
<p><strong>Input:</strong> <span class="example-io">s = "accca", 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>"acbca"</code>.</p>
|
|
|
|
<p>Then we perform the operations:</p>
|
|
|
|
<ol>
|
|
<li>The longest prefix containing at most 2 distinct characters is <code>"ac"</code>, we remove it and <code>s</code> becomes <code>"bca"</code>.</li>
|
|
<li>Now The longest prefix containing at most 2 distinct characters is <code>"bc"</code>, so we remove it and <code>s</code> becomes <code>"a"</code>.</li>
|
|
<li>Finally, we remove <code>"a"</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 = "aabaab", k = 3</span></p>
|
|
|
|
<p><strong>Output:</strong> <span class="example-io">1</span></p>
|
|
|
|
<p><strong>Explanation:</strong></p>
|
|
|
|
<p>Initially <code>s</code> 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 = "xxyz", 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 <code>s[0]</code> or <code>s[1]</code> to something other than characters in <code>s</code>, for example, to change <code>s[0]</code> to <code>w</code>.</p>
|
|
|
|
<p>Then <code>s</code> becomes <code>"wxyz"</code>, which consists of 4 distinct characters, so as <code>k</code> is 1, it will divide into 4 partitions.</p>
|
|
</div>
|
|
|
|
<p> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
|
|
<li><code>s</code> consists only of lowercase English letters.</li>
|
|
<li><code>1 <= k <= 26</code></li>
|
|
</ul>
|