You are given a string s and an integer k.
In one operation, you can replace the character at any position with the next or previous letter in the alphabet (wrapping around so that 'a' is after 'z'). For example, replacing 'a' with the next letter results in 'b', and replacing 'a' with the previous letter results in 'z'. Similarly, replacing 'z' with the next letter results in 'a', and replacing 'z' with the previous letter results in 'y'.
Return the length of the longest palindromic subsequence of s that can be obtained after performing at most k operations.
\n
Example 1:
\n\nInput: s = "abced", k = 2
\n\nOutput: 3
\n\nExplanation:
\n\ns[1] with the next letter, and s becomes "acced".s[4] with the previous letter, and s becomes "accec".The subsequence "ccc" forms a palindrome of length 3, which is the maximum.
Example 2:
\n\nInput: s = "aaazzz", k = 4
\n\nOutput: 6
\n\nExplanation:
\n\ns[0] with the previous letter, and s becomes "zaazzz".s[4] with the next letter, and s becomes "zaazaz".s[3] with the next letter, and s becomes "zaaaaz".The entire string forms a palindrome of length 6.
\n\n
Constraints:
\n\n1 <= s.length <= 2001 <= k <= 200s consists of only lowercase English letters.给你一个字符串 s 和一个整数 k。
在一次操作中，你可以将任意位置的字符替换为字母表中相邻的字符（字母表是循环的，因此 'z' 的下一个字母是 'a'）。例如，将 'a' 替换为下一个字母结果是 'b'，将 'a' 替换为上一个字母结果是 'z'；同样，将 'z' 替换为下一个字母结果是 'a'，替换为上一个字母结果是 'y'。
返回在进行 最多 k 次操作后，s 的 最长回文子序列 的长度。
子序列 是一个 非空 字符串，可以通过删除原字符串中的某些字符（或不删除任何字符）并保持剩余字符的相对顺序得到。
\n\n回文 是正着读和反着读都相同的字符串。
\n\n\n\n
示例 1：
\n\n输入: s = \"abced\", k = 2
\n\n输出: 3
\n\n解释:
\n\ns[1] 替换为下一个字母，得到 \"acced\"。s[4] 替换为上一个字母，得到 \"accec\"。子序列 \"ccc\" 形成一个长度为 3 的回文，这是最长的回文子序列。
示例 2：
\n\n输入: s = \"aaazzz\", k = 4
\n\n输出: 6
\n\n解释:
\n\ns[0] 替换为上一个字母，得到 \"zaazzz\"。s[4] 替换为下一个字母，得到 \"zaazaz\"。s[3] 替换为下一个字母，得到 \"zaaaaz\"。整个字符串形成一个长度为 6 的回文。
\n\n\n
提示:
\n\n1 <= s.length <= 2001 <= k <= 200s 仅由小写英文字母组成。dp[i][j][k] is the length of the longest palindromic subsequence in substring [i..j] with cost at most k.",
"dp[i][j][k] = max(dp[i + 1][j][k], dp[i][j - 1][k], dp[i + 1][j - 1][k - dist(s[i], s[j])] + 2), where dist(x, y) is the minimum cyclic distance between x and y."
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"sampleTestCase": "\"abced\"\n2",
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