take-k-of-each-character-from-left-and-right

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

You are given a string <code>s</code> consisting of the characters <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, and <code>&#39;c&#39;</code> and a non-negative integer <code>k</code>. Each minute, you may take either the leftmost character of <code>s</code>, or the rightmost character of <code>s</code>.

Return the minimum number of minutes needed for you to take at least <code>k</code> of each character, or return <code>-1</code> if it is not possible to take <code>k</code> of each character.

&nbsp;
Example 1:

<pre>
Input: s = &quot;aabaaaacaabc&quot;, k = 2
Output: 8
Explanation: 
Take three characters from the left of s. You now have two &#39;a&#39; characters, and one &#39;b&#39; character.
Take five characters from the right of s. You now have four &#39;a&#39; characters, two &#39;b&#39; characters, and two &#39;c&#39; characters.
A total of 3 + 5 = 8 minutes is needed.
It can be proven that 8 is the minimum number of minutes needed.
</pre>

Example 2:

<pre>
Input: s = &quot;a&quot;, k = 1
Output: -1
Explanation: It is not possible to take one &#39;b&#39; or &#39;c&#39; so return -1.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 105</code></li>
	<li><code>s</code> consists of only the letters <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, and <code>&#39;c&#39;</code>.</li>
	<li><code>0 &lt;= k &lt;= s.length</code></li>
</ul>

takeCharacters

Take K of Each Character From Left and Right

Merge Sorted Array

merge-sorted-array

合并两个有序数组

Reorder List

reorder-list

重排链表

Defuse the Bomb

defuse-the-bomb

拆炸弹

Hash Table

String

Sliding Window

给你一个由字符 <code>'a'</code>、<code>'b'</code>、<code>'c'</code> 组成的字符串 <code>s</code> 和一个非负整数 <code>k</code> 。每分钟，你可以选择取走 <code>s</code> 最左侧 还是 最右侧 的那个字符。

你必须取走每种字符 至少 <code>k</code> 个，返回需要的 最少 分钟数；如果无法取到，则返回 <code>-1</code> 。

&nbsp;

示例 1：

<pre>
输入：s = "aabaaaacaabc", k = 2
输出：8
解释：
从 s 的左侧取三个字符，现在共取到两个字符 'a' 、一个字符 'b' 。
从 s 的右侧取五个字符，现在共取到四个字符 'a' 、两个字符 'b' 和两个字符 'c' 。
共需要 3 + 5 = 8 分钟。
可以证明需要的最少分钟数是 8 。
</pre>

示例 2：

<pre>
输入：s = "a", k = 1
输出：-1
解释：无法取到一个字符 'b' 或者 'c'，所以返回 -1 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 105</code></li>
	<li><code>s</code> 仅由字母 <code>'a'</code>、<code>'b'</code>、<code>'c'</code> 组成</li>
	<li><code>0 &lt;= k &lt;= s.length</code></li>
</ul>