You are given a string s and an integer k.
First, you are allowed to change at most one index in s to another lowercase English letter.
After that, do the following partitioning operation until s is empty:
s containing at most k distinct characters.s and increase the number of partitions by one. The remaining characters (if any) in s maintain their initial order.Return an integer denoting the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.
\n\n\n
Example 1:
\n\nInput: s = "accca", k = 2
\n\nOutput: 3
\n\nExplanation:
\n\nThe optimal way is to change s[2] to something other than a and c, for example, b. then it becomes "acbca".
Then we perform the operations:
\n\n"ac", we remove it and s becomes "bca"."bc", so we remove it and s becomes "a"."a" and s becomes empty, so the procedure ends.Doing the operations, the string is divided into 3 partitions, so the answer is 3.
\nExample 2:
\n\nInput: s = "aabaab", k = 3
\n\nOutput: 1
\n\nExplanation:
\n\nInitially s contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.
Example 3:
\n\nInput: s = "xxyz", k = 1
\n\nOutput: 4
\n\nExplanation:
\n\nThe optimal way is to change s[0] or s[1] to something other than characters in s, for example, to change s[0] to w.
Then s becomes "wxyz", which consists of 4 distinct characters, so as k is 1, it will divide into 4 partitions.
\n
Constraints:
\n\n1 <= s.length <= 104s consists only of lowercase English letters.1 <= k <= 26给你一个下标从 0 开始的字符串 s 和一个整数 k。
你需要执行以下分割操作，直到字符串 s 变为 空：
s 的最长 前缀，该前缀最多包含 k 个 不同 字符。s 中保持原来的顺序。执行操作之 前 ，你可以将 s 中 至多一处 下标的对应字符更改为另一个小写英文字母。
在最优选择情形下改变至多一处下标对应字符后，用整数表示并返回操作结束时得到的 最大 分割数量。
\n\n\n\n
示例 1：
\n\n输入：s = \"accca\", k = 2
\n\n输出：3
\n\n解释：
\n\n最好的方式是把 s[2] 变为除了 a 和 c 之外的东西，比如 b。然后它变成了 \"acbca\"。
然后我们执行以下操作：
\n\n\"ac\"，我们删除它然后 s 变为 \"bca\"。\"bc\"，所以我们删除它然后 s 变为 \"a\"。\"a\" 并且 s 变成空串，所以该过程结束。进行操作时，字符串被分成 3 个部分，所以答案是 3。
\n示例 2：
\n\n输入：s = \"aabaab\", k = 3
\n\n输出：1
\n\n解释：
\n\n一开始 s 包含 2 个不同的字符，所以无论我们改变哪个， 它最多包含 3 个不同字符，因此最多包含 3 个不同字符的最长前缀始终是所有字符，因此答案是 1。
示例 3：
\n\n输入：s = \"xxyz\", k = 1
\n\n输出：4
\n\n解释：
\n\n最好的方式是将 s[0] 或 s[1] 变为 s 中字符以外的东西，例如将 s[0] 变为 w。
然后 s 变为 \"wxyz\"，包含 4 个不同的字符，所以当 k 为 1，它将分为 4 个部分。
\n\n
提示：
\n\n1 <= s.length <= 104s 只包含小写英文字母。1 <= k <= 26pref[i]: The number of resulting partitions from the operations by performing the operations on s[0:i].suff[i]: The number of resulting partitions from the operations by performing the operations on s[i:n - 1], where n == s.length.partition_start[i]: The start index of the partition containing the ith index after performing the operations.i, we can try all possible 25 replacements:r, such that the number of distinct characters in the range [partition_start[i], r] is at most k.2 cases:r >= i: the number of resulting partitions in this case is 1 + pref[partition_start[i] - 1] + suff[r + 1].r2 such that the number of distinct characters in the range [r:r2] is at most k. 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