3003. Maximize the Number of Partitions After Operations

Problem Link

Description

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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:

• Choose the longest prefix of s containing at most k distinct characters.
• Delete the prefix from 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:

1. The longest prefix containing at most 2 distinct characters is "ac", we remove it and s becomes "bca".
2. Now The longest prefix containing at most 2 distinct characters is "bc", so we remove it and s becomes "a".
3. Finally, we remove "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 <= 104
• s consists only of lowercase English letters.
• 1 <= k <= 26

Solution

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Python3

class Solution:
    def maxPartitionsAfterOperations(self, s: str, k: int) -> int:
        N = len(s)
        s = [ord(x) - ord('a') for x in s]
        
        @cache
        def go(index, left, mask):
            if index == N: return 0
            
            newMask = mask | (1 << s[index])
            bitCount = newMask.bit_count()
            
            if bitCount > k:
                res = 1 + go(index + 1, left, 1 << s[index])
            else:
                res = go(index + 1, left, newMask)
            
            if left > 0:
                for c in range(26):
                    newMask = mask | (1 << c)
                    bitCount = newMask.bit_count()
                    
                    if bitCount > k:
                        res = max(res, 1 + go(index + 1, left - 1, 1 << c))
                    else:
                        res = max(res, go(index + 1, left - 1, newMask))
                    
            return res
        
        return go(0, 1, 0) + 1