#3003

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Maximize the Number of Partitions After Operations

Maximizing the number of partitions in a string after changing one character and applying partitioning operations using dynamic programming.

string dynamic programming bit manipulation bitmask

Problem Statement

You are given a string s and an integer k. You can change at most one character in the string to any other lowercase English letter. After that, repeatedly partition the string such that each part contains at most k distinct characters. The task is to maximize the number of partitions possible.

To solve this, you need to explore different character replacements and partitioning strategies. The challenge is to determine the best partitioning by considering all possible changes and how they impact the string's distinct character count during the partitioning process.

Examples

Example 1

Input: s = "accca", k = 2

Output: 3

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: Doing the operations, the string is divided into 3 partitions, so the answer is 3.

Example 2

Input: s = "aabaab", k = 3

Output: 1

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

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 Approach

State Transition Dynamic Programming

This approach uses dynamic programming to track the number of partitions possible for each substring, while considering the replacement of a single character at each position in the string. The state transitions help compute the maximum partitions while optimizing for the allowed number of distinct characters.

Brute-Force Character Replacements

For each position in the string, try brute-forcing a replacement with all other characters. This allows you to explore how the replacement impacts the string's distinct character set and, consequently, the number of partitions that can be formed.

Optimizing with Bit Manipulation

Bit manipulation can be employed to efficiently track the distinct characters in each substring. This reduces the time complexity when checking possible partitioning options and enables quicker state transitions during dynamic programming steps.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach, considering both the number of positions in the string and the character replacements being brute-forced. Space complexity is influenced by the need to store the states of different substrings and their partitioning results.

What Interviewers Usually Probe

Look for the candidate's understanding of state transitions in dynamic programming.
Evaluate their approach to managing distinct characters using bit manipulation.
Assess their ability to brute-force replacements and handle various string configurations.

Common Pitfalls or Variants

Common pitfalls

Overlooking the impact of character replacements on the distinct character count during partitioning.
Failing to account for the optimal replacement at each index, resulting in fewer partitions.
Mismanaging the dynamic programming states and transitions, leading to inefficient solutions.

Follow-up variants

Maximizing partitions with more than one allowed replacement.
Limiting the number of partitions instead of maximizing them.
Changing characters at multiple indices to optimize partitioning.

FAQ

How do I approach the 'Maximize the Number of Partitions After Operations' problem?

Start by understanding how state transitions work in dynamic programming. Then, try brute-forcing character replacements at each index while considering the allowed distinct character limit during partitioning.

What dynamic programming techniques can I use for this problem?

Use dynamic programming to store the best partitioning results for each substring, taking into account character replacements and the allowed number of distinct characters.

What is the role of bit manipulation in this problem?

Bit manipulation is used to efficiently track the distinct characters in substrings, speeding up the process of checking possible partitions during dynamic programming steps.

Can the number of partitions be increased by changing more than one character?

The problem specifically allows only one character replacement, so you must maximize partitions within this constraint. Changing more than one character would alter the problem's requirements.

How do I handle edge cases with minimal distinct characters?

In cases where the string already has fewer distinct characters than k, you should focus on maximizing partitions by considering the best character replacement strategy for the optimal partitioning.

terminal

Solution

Solution 1: Memoized Search

We design a function $\textit{dfs}(i, \textit{cur}, t)$ that represents the maximum number of partitions we can obtain when currently processing index $i$ of string $s$, the current prefix already contains the character set $\textit{cur}$, and we can still modify $t$ characters. Then the answer is $\textit{dfs}(0, 0, 1)$.

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class Solution:
    def maxPartitionsAfterOperations(self, s: str, k: int) -> int:
        @cache
        def dfs(i: int, cur: int, t: int) -> int:
            if i >= n:
                return 1
            v = 1 << (ord(s[i]) - ord("a"))
            nxt = cur | v
            if nxt.bit_count() > k:
                ans = dfs(i + 1, v, t) + 1
            else:
                ans = dfs(i + 1, nxt, t)
            if t:
                for j in range(26):
                    nxt = cur | (1 << j)
                    if nxt.bit_count() > k:
                        ans = max(ans, dfs(i + 1, 1 << j, 0) + 1)
                    else:
                        ans = max(ans, dfs(i + 1, nxt, 0))
            return ans

        n = len(s)
        return dfs(0, 0, 1)

Continue Practicing

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