Watch 10 video solutions for Maximize the Number of Partitions After Operations, a hard level problem involving String, Dynamic Programming, Bit Manipulation. This walkthrough by NeetCodeIO has 101,495 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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.
Example 1:
Input: s = "accca", k = 2
Output: 3
Explanation:
The 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:
"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.
Example 2:
Input: s = "aabaab", k = 3
Output: 1
Explanation:
Initially 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:
Input: s = "xxyz", k = 1
Output: 4
Explanation:
The 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.
Constraints:
1 <= s.length <= 104s consists only of lowercase English letters.1 <= k <= 26