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Maximize the Number of Partitions After Operations

Difficulty: Hard

Problem Description

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, perform a partitioning operation until s is empty by choosing the longest prefix of s containing at most k distinct characters, deleting that prefix, and increasing the number of partitions by one. Return the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.

Key Insights

  • The problem requires maximizing the number of partitions by changing one character in the string.
  • To determine the longest prefix with at most k distinct characters, a sliding window approach can be utilized.
  • The distinct character count can change based on the character chosen for replacement, allowing for different partitioning results.

Space and Time Complexity

Time Complexity: O(n * 26) where n is the length of the string (up to 10^4) and 26 represents the possible lowercase letters to replace. Space Complexity: O(1) for constant space used for counting distinct characters.

Solution

The solution involves iterating through each character in the string as a potential character to change. For each character, we replace it with every possible lowercase letter (from 'a' to 'z') that is not in the current substring. We then use a sliding window technique to find the longest prefix that contains at most k distinct characters. This process is repeated for each character in the string to determine the maximum number of partitions.

Code Solutions

def maxPartitions(s: str, k: int) -> int:
    def longest_prefix_with_k_distinct(s, k):
        char_count = {}
        left = 0
        max_length = 0
        
        for right in range(len(s)):
            char_count[s[right]] = char_count.get(s[right], 0) + 1
            
            while len(char_count) > k:
                char_count[s[left]] -= 1
                if char_count[s[left]] == 0:
                    del char_count[s[left]]
                left += 1
            
            max_length = max(max_length, right - left + 1)
        
        return max_length

    max_partitions = 0
    original_distinct_count = len(set(s))

    for i in range(len(s)):
        for c in range(26):
            new_char = chr(c + ord('a'))
            if new_char != s[i]:
                modified_s = s[:i] + new_char + s[i + 1:]
                max_partitions = max(max_partitions, longest_prefix_with_k_distinct(modified_s, k))
    
    if original_distinct_count <= k:
        return 1
    
    return max_partitions
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