minimum-number-of-moves-to-make-palindrome

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Algorithms

You are given a string <code>s</code> consisting only of lowercase English letters.

In one move, you can select any two adjacent characters of <code>s</code> and swap them.

Return the minimum number of moves needed to make <code>s</code> a palindrome.

Note that the input will be generated such that <code>s</code> can always be converted to a palindrome.

&nbsp;
Example 1:

<pre>
Input: s = &quot;aabb&quot;
Output: 2
Explanation:
We can obtain two palindromes from s, &quot;abba&quot; and &quot;baab&quot;. 
- We can obtain &quot;abba&quot; from s in 2 moves: &quot;aabb&quot; -&gt; &quot;abab&quot; -&gt; &quot;abba&quot;.
- We can obtain &quot;baab&quot; from s in 2 moves: &quot;aabb&quot; -&gt; &quot;abab&quot; -&gt; &quot;baab&quot;.
Thus, the minimum number of moves needed to make s a palindrome is 2.
</pre>

Example 2:

<pre>
Input: s = &quot;letelt&quot;
Output: 2
Explanation:
One of the palindromes we can obtain from s in 2 moves is &quot;lettel&quot;.
One of the ways we can obtain it is &quot;letelt&quot; -&gt; &quot;letetl&quot; -&gt; &quot;lettel&quot;.
Other palindromes such as &quot;tleelt&quot; can also be obtained in 2 moves.
It can be shown that it is not possible to obtain a palindrome in less than 2 moves.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
	<li><code>s</code> can be converted to a palindrome using a finite number of moves.</li>
</ul>

minMovesToMakePalindrome

Minimum Number of Moves to Make Palindrome

Minimum Insertion Steps to Make a String Palindrome

minimum-insertion-steps-to-make-a-string-palindrome

让字符串成为回文串的最少插入次数

Minimum Number of Flips to Make Binary Grid Palindromic I

minimum-number-of-flips-to-make-binary-grid-palindromic-i

最少翻转次数使二进制矩阵回文 I

Greedy

Binary Indexed Tree

Two Pointers

String

给你一个只包含小写英文字母的字符串&nbsp;<code>s</code>&nbsp;。

每一次 操作&nbsp;，你可以选择 <code>s</code>&nbsp;中两个 相邻&nbsp;的字符，并将它们交换。

请你返回将 <code>s</code>&nbsp;变成回文串的 最少操作次数&nbsp;。

注意&nbsp;，输入数据会确保&nbsp;<code>s</code>&nbsp;一定能变成一个回文串。

&nbsp;

示例 1：

<pre>输入：s = "aabb"
输出：2
解释：
我们可以将 s 变成 2 个回文串，"abba" 和 "baab" 。
- 我们可以通过 2 次操作得到 "abba" ："aabb" -&gt; "abab" -&gt; "abba" 。
- 我们可以通过 2 次操作得到 "baab" ："aabb" -&gt; "abab" -&gt; "baab" 。
因此，得到回文串的最少总操作次数为 2 。
</pre>

示例 2：

<pre>输入：s = "letelt"
输出：2
解释：
通过 2 次操作从 s 能得到回文串 "lettel" 。
其中一种方法是："letelt" -&gt; "letetl" -&gt; "lettel" 。
其他回文串比方说 "tleelt" 也可以通过 2 次操作得到。
可以证明少于 2 次操作，无法得到回文串。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code>&nbsp;只包含小写英文字母。</li>
	<li><code>s</code>&nbsp;可以通过有限次操作得到一个回文串。</li>
</ul>