You are given a string s consisting of lowercase English letters, and you are allowed to perform operations on it. In one operation, you can replace a character in s with another lowercase English letter.
Your task is to make s a palindrome with the minimum number of operations possible. If there are multiple palindromes that can be made using the minimum number of operations, make the lexicographically smallest one.
A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Return the resulting palindrome string.
\n\n\n
Example 1:
\n\n\nInput: s = "egcfe"\nOutput: "efcfe"\nExplanation: The minimum number of operations to make "egcfe" a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is "efcfe", by changing 'g'.\n\n\n
Example 2:
\n\n\nInput: s = "abcd"\nOutput: "abba"\nExplanation: The minimum number of operations to make "abcd" a palindrome is 2, and the lexicographically smallest palindrome string we can get by modifying two characters is "abba".\n\n\n
Example 3:
\n\n\nInput: s = "seven"\nOutput: "neven"\nExplanation: The minimum number of operations to make "seven" a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is "neven".\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 1000s consists of only lowercase English letters.给你一个由 小写英文字母 组成的字符串 s ，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 s 中的一个字符。
请你执行 尽可能少的操作 ，使 s 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 a 和 b ，在 a 和 b 出现不同的第一个位置，如果该位置上 a 中对应字母比 b 中对应字母在字母表中出现顺序更早，则认为 a 的字典序比 b 的字典序要小。
返回最终的回文字符串。
\n\n\n\n
示例 1：
\n\n\n输入：s = \"egcfe\"\n输出：\"efcfe\"\n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' 。\n\n\n
示例 2：
\n\n\n输入：s = \"abcd\"\n输出：\"abba\"\n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\" 。\n\n\n
示例 3：
\n\n\n输入：s = \"seven\"\n输出：\"neven\"\n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\" 。\n\n
\n\n
提示：
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collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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