lexicographically-smallest-palindrome

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Algorithms

You are given a string <code node="[object Object]">s</code> consisting of lowercase English letters, and you are allowed to perform operations on it. In one operation, you can replace a character in <code node="[object Object]">s</code> with another lowercase English letter.

Your task is to make <code node="[object Object]">s</code> a palindrome with the minimum number of operations possible. If there are multiple palindromes that can be <meta charset="utf-8" />made using the minimum number of operations, <meta charset="utf-8" />make the lexicographically smallest one.

A string <code>a</code> is lexicographically smaller than a string <code>b</code> (of the same length) if in the first position where <code>a</code> and <code>b</code> differ, string <code>a</code> has a letter that appears earlier in the alphabet than the corresponding letter in <code>b</code>.

Return the resulting palindrome string.

&nbsp;
Example 1:

<pre>
Input: s = &quot;egcfe&quot;
Output: &quot;efcfe&quot;
Explanation: The minimum number of operations to make &quot;egcfe&quot; a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is &quot;efcfe&quot;, by changing &#39;g&#39;.
</pre>

Example 2:

<pre>
Input: s = &quot;abcd&quot;
Output: &quot;abba&quot;
Explanation: The minimum number of operations to make &quot;abcd&quot; a palindrome is 2, and the lexicographically smallest palindrome string we can get by modifying two characters is &quot;abba&quot;.
</pre>

Example 3:

<pre>
Input: s = &quot;seven&quot;
Output: &quot;neven&quot;
Explanation: The minimum number of operations to make &quot;seven&quot; a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is &quot;neven&quot;.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code>&nbsp;consists of only lowercase English letters.</li>
</ul>

makeSmallestPalindrome

Lexicographically Smallest Palindrome

Greedy

Two Pointers

String

给你一个由 小写英文字母 组成的字符串 <code>s</code> ，你可以对其执行一些操作。在一步操作中，你可以用其他小写英文字母 替换&nbsp; <code>s</code> 中的一个字符。

请你执行 尽可能少的操作 ，使 <code>s</code> 变成一个 回文串 。如果执行 最少 操作次数的方案不止一种，则只需选取 字典序最小 的方案。

对于两个长度相同的字符串 <code>a</code> 和 <code>b</code> ，在 <code>a</code> 和 <code>b</code> 出现不同的第一个位置，如果该位置上 <code>a</code> 中对应字母比 <code>b</code> 中对应字母在字母表中出现顺序更早，则认为 <code>a</code> 的字典序比 <code>b</code> 的字典序要小。

返回最终的回文字符串。

&nbsp;

示例 1：

<pre>
输入：s = "egcfe"
输出："efcfe"
解释：将 "egcfe" 变成回文字符串的最小操作次数为 1 ，修改 1 次得到的字典序最小回文字符串是 "efcfe"，只需将 'g' 改为 'f' 。
</pre>

示例 2：

<pre>
输入：s = "abcd"
输出："abba"
解释：将 "abcd" 变成回文字符串的最小操作次数为 2 ，修改 2 次得到的字典序最小回文字符串是 "abba" 。
</pre>

示例 3：

<pre>
输入：s = "seven"
输出："neven"
解释：将 "seven" 变成回文字符串的最小操作次数为 1 ，修改 1 次得到的字典序最小回文字符串是 "neven" 。</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> 仅由小写英文字母组成</li>
</ul>